Identify m∠EFH.

The figure shows triangle G F H. Angle G is equal to 76 degrees. A measure of angle H is given by the formula 3 times x minus 3 degrees. Side G F lies on ray G F with endpoint G. Angle H F E is an exterior angle and its measure is given by the formula 8 times x plus 8 degrees.

Answer:

1921.68 [tex]ft^{3}[/tex]

Step-by-step explanation:

V = pi*r*r*h

= 3.14*6*6*17

= 1921.68

The correct proportions are given as follows:

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

The proportions for this problem are given dividing y by x, hence:

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Answer: 3n ants

Step-by-step explanation:

         If there are n ants one month, and the ants triple every month, then next month there will be 3n ants.

         We can multiply 3 by n to show it tripling.

      To test this, 3(10,000) = 30,000 and 3(270,000) = 810,000, which corresponds to what you've filled out in the table.

Answer:

I don't understand what the question is what do you need to find?

The points on the graph of (-4, 0), (-2.5, -12), and (0, -3), gives;

From the description of the graph, we have;

Furthest point left of the graph = (-4, 0)

The furthest point right on the graph = (0, -3) = The maximum point

The minimum point = (-2.5, -12)

F(x) < 0 at the minimum point

The minimum point is to the right of x = -4

The point the graph crosses the y-axis = (0, -3)

Therefore;

The interval of the graph where F(x) is larger than 0 is to the left of (-4, 0), is the interval (-∞, -4)

The true statement is therefore;

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Answer:

-0.4 or [tex]\frac{-2}{ 5}[/tex]

Step-by-step explanation:

[tex]tan^-^{1} ({\frac{2}{5} })=21.80\\[/tex]

tan ( 2π-21.80)

but 2π = 360

Therefore our question becomes:

tan ( 360-21.80)=tan(338.2)= -0.4

Question 11

1) [tex]\overline{BA} \cong \overline{FA}[/tex], [tex]\angle 1 \cong \angle 2[/tex] (given)

2) [tex]\angle A \cong \angle A[/tex] (reflexive property)

3) [tex]\triangle AEB \cong \triangle ACF[/tex] (ASA)

4) [tex]\overline{AC} \cong \overline{AE}[/tex] (CPCTC)

Question 12

1) Isosceles [tex]\triangle ACD[/tex] with [tex]\overline{AC} \cong \overline{AD}[/tex], [tex]\overline{BC} \cong \overline{ED}[/tex] (given)

2) [tex]\angle ACD \cong \angle ADC[/tex] (angles opposite congruent sides in a triangle are congruent)

3) [tex]\angle ACB[/tex] and [tex]\angle ACD[/tex] are supplementary. [tex]\angle ADC[/tex] and [tex]\angle ADE[/tex] are supplementary (angles that form a linear pair are supplementary)

4) [tex]\angle ACB \cong \angle ADE[/tex] (supplements of congruent angles are congruent)

5) [tex]\triangle ABC \cong \triangle AED[/tex] (SAS)

6) [tex]\overline{AB} \cong \overline{AE}[/tex] (CPCTC)

7) [tex]\triangle ABE[/tex] is an isosceles triangle (a triangle with two congruent sides is isosceles)

Note: I changed the names of the segments in Question 11 because of the word filter.

The missing statements that completes the proof that proves that quadrilateral ACDB is a parallelogram is explained below.

A quadrilateral that has the following properties is a parallelogram, in reference to the quadrilateral ACDB in the image given:

Given that ΔCFD ≅ ΔBGA, the proof that quadrilateral ACDB is a parallelogram is given below:

Since we know that ΔCFD ≅ ΔBGA, therefore, CD ≅ BA because corresponding parts of congruent triangles are: congruent.

Since we also know that m∠DCA + m∠CAB = 180°, then we also know that both angles are: supplementary.

Transversal A.F cut the two lines, CD and BA, then the interior angles on the same side of the transversal are: supplementary, thus, we it implies we know that lines CD and BA would be: parallel.

Since one pair of sides (CD and BA) are both parallel and congruent, then we know that quadrilateral ACDB is a parallelogram.

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[tex]y = 9x {}^{2} - 4 \\ flip \: x \: and \: y \\ x =9 (y ) {}^{2} - 4 \\ solve \: for \: the \: inverse \: of \: y \\ x + 4 = 9(y ) {}^{2} \\ y {} {}^{2} = \frac{x + 4}{9} \\ you \: can \: split \: it \: into \: two[/tex]

[tex]y {}^{(1)} = \sqrt{ \frac{x + 4}{9} } \\ y {}^{(2)} = - \sqrt{ \frac{x + 4}{9} } [/tex]

Step-by-step explanation:

as the graphic shows, there are 2 objects : 1 block and 1 triangular shaped half-block.

the block is

7cm × 6cm × 2cm

the half-block is

8cm × 7cm × 6cm

with 10cm being the length of the tilted "roof" area.

the 2 sides facing each other are not visible to the observer, so, they are not part of the surface area of the composite figure.

let's start with the block :

top and bottom 7×2

front and back 6×2

no left (fully covered by the half-block)

right 6×7

that gives us :

2 × 7×2 = 2×14 = 28 cm²

2 × 6×2 = 2×12 = 24 cm²

6×7 = 42 cm²

in total : 94 cm²

the half-block :

top 10×7

bottom 8×7

front and back (triangles) 8×6/2

no left (due to being a half-block)

no right (fully covered by the block)

that gives us :

10×7 = 70 cm²

8×7 = 56 cm²

2 × 8×6/2 = 2×24 = 48 cm²

in total : 174 cm²

so, the complete surface area of the composite figure is

94 + 174 = 268 cm²

Area of the entire rectangle is x² + 11x + 30. This can be obtained by adding area of each square.

Method 1

Area of rectangle = length × width  

Area of entire rectangle = x² + 5x + 6x + 30 = x² + 11x + 30

Method 2

Area of entire rectangle = length × width

                                   = (x+6)×(x+5)    

                                  = x² + (6+5)x + (6×5)        [∵(x+a)(x+b) =x² + (a+b)x + ab]

                                    = x² + 11x + 30

Hence area of the entire rectangle is x² + 11x + 30.

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Using the probability concept, the correct statement is:

B. P(hibiscus|red) = P(hibiscus).

A probability is given by the number of desired outcomes divided by the number of total outcomes.

For item a, the probabilities are:

Same probabilities, hence the statement that they are different is false.

For item b, the probabilities are:

Equal, hence this is the correct statement.

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Answer:

[tex]\textsf{1.} \quad C^{-1}(F)=\dfrac{9}{5}F+32[/tex]

2.   [-273.15, ∞)

Step-by-step explanation:

Given:

[tex]C=\dfrac{5}{9}(F-32), \quad \text{where }F \geq -459.67[/tex]

To find the inverse of the given function, make F the subject:

[tex]\begin{aligned}C & =\dfrac{5}{9}(F-32)\\\implies \dfrac{9}{5}C & =F-32\\\implies F & = \dfrac{9}{5}C+32 \end{aligned}[/tex]

[tex]\textsf{Replace the } F \textsf{ with }C^{-1}(F)\textsf{ and the }C \textsf{ with } F:[/tex] :

[tex]\implies C^{-1}(F)=\dfrac{9}{5}F+32[/tex]

Domain: set of all possible input values (x-values)

Range: set of all possible output values (y-values)

The given domain of the function C(F) is F ≥ -459.67

Therefore, the minimum value of the function is:

[tex]\implies C(-459.67)=\dffrac{5}{9}(-459.67-32)=-273.15[/tex]

This means the range of the function C(F) is C(F) ≥ -273.15

The domain of the inverse function is the range of the function.

Therefore, the domain of the inverse function in interval notation is:  [-273.15, ∞)

Answer: -273.15

Step-by-step explanation:

The solution set to the inequality 5.1(3 + 2.2x) > –14.25 – 6(1.7x + 4) gives x > -2.5

An equation is an expression that shows the relationship between two numbers and variables.

An independent variable is a variable that does not depend on any other variable for its value whereas a dependent variable is a variable that depend on any other variable for its value.

Inequality shows the non equal comparison of two or more numbers and variables.

Given the inequality:

5.1(3 + 2.2x) > –14.25 – 6(1.7x + 4)

Simplifying:

15.3 + 11.22x > –14.25 – 10.2x - 24

21.42x > -53.55

x > -2.5

The solution set to the inequality 5.1(3 + 2.2x) > –14.25 – 6(1.7x + 4) gives x > -2.5

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Large floods in 2007 and 2010 changed the flood recurrence curve.

What caused the Red River flood 1997?

A extremely unusual thawing of winter snow and river ice after a winter season that saw far above-normal precipitation across the Northern Plains was the main contributor to the flooding in 1997.

Why is the Red River called the Red River?

After it was explored in 1732–33 by the French voyageur Pierre Gaultier de Varennes et de La Vérendrye, the river, called Red because of the reddish brown silt it carries, served as a transportation link between Lake Winnipeg and the Mississippi River system.

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Using the normal distribution, we have that:

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The mean and the standard deviation for this problem are given, respectively, by:

[tex]\mu = 5.34, \sigma = 0.03[/tex]

The 8th percentile separates the bottom 8%, that is, X when Z = -1.405, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.405 = \frac{X - 5.34}{0.03}[/tex]

X - 5.34 = -1.405 x 0.03

X = 5.30.

The 92th percentile separates the top 8%, that is, X when Z = 1.405, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.405 = \frac{X - 5.34}{0.03}[/tex]

X - 5.34 = 1.405 x 0.03

X = 5.38.

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The number to add to the set of data is 26

The set of data is given as:

11,14,23 and 16

Let the new number be x

So that

Mean = 18

Mean is calculated as:

Mean = Sum/Count

So, we have:

(11 + 14 + 23 + 16 + x)/5 = 18

Multiply by 5

11 + 14 + 23 + 16 + x = 90

Evaluate the like terms

x = 26

Hence, the number to add is 26

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Complete question

A set of data already has the values 11, 14, 23, and 16. What value would have to be added to the set for the mean of the five numbers to be 18?

The area of the plate covered by the napkin is (C) [tex]56cm^{2}[/tex].

What is an area of a triangle?

To find the area of the plate covered by the napkin:

Given :

The following steps can be used in order to determine the area of the plate covered by the napkin:

Step 1 - The formula of the perimeter of a triangle is given below:

P = a + b + c   --- (1)

where a, b, and c are the length of the sides of the triangle.

Step 2 - According to the given data, the triangle is isosceles.

Therefore, the two sides are similar and the length of the base is 8 cm.

Step 3 - Now, substitute the values of the known terms in the above formula.

38 = 2L + 8

38 - 8 = 2L

L = 15 cm

Step 4 - The area of the plate covered by the napkin is given below:

[tex]A=\frac{1}{2} * 15 * 15 *Sin30[/tex]

Step 5 - Simplify the above expression.

[tex]A=56.25cm^{2}[/tex]

Rounding off to the nearest whole number, which is [tex]56cm^{2}[/tex].

Therefore, the area of the plate covered by the napkin is (C) [tex]56cm^{2}[/tex].

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The question you are looking for is here:

A napkin is folded into an isosceles triangle, triangle ABC, and placed on a plate, as shown. The napkin has a perimeter of 38 centimeters. To the nearest square centimeter, how many square centimeters of the plate is covered by the napkin?

(A) 16 square centimeters

(B) 30 square centimeters

(C) 56 square centimeters

(D) 60 square centimeters

27a³ + 189a²b + 441ab² + 343b³ = (3a + 7b)³

27a³ + 189a²b + 441ab² + 343b³

Therefore,

27a³  = (3a)³

189a²b  = 3(3a)²(7b)

441ab² = 3(3a)(7b)²

343b³ = (7b)³

(3a)³ + (7b)³ + 3(3a)²(7b) + 3(3a)(7b)²

a³ + b³ + 3a²b + 3ab²  = (a + b)³

Therefore,

27a³ + 189a²b + 441ab² + 343b³ = (3a + 7b)³

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The values which completes the table regarding summer camp is option b which is a=15,b=7, c=5, d=10, e=12.

Given that there are 32 campers, 22 of them can swim, 20 play softball and 5 do not play softball or swim.

We have to find the values of a,b,c,d,e so that we can complete the table.

Table is a combination of rows and columns. In our case the third row and third column shows the total.

from the table we can write that 22+d=32-----------1

so d=32-22

=10

d=10

c+5=d-----------2

c=10-5=5

c=5

a+c=20------------------3

a+5=20

a=20-5

a=15

a+b=22--------------3

15+b=22

b=7

20+e=32----------4

e=32-22

e=10.

Hence the values which completes the table is a=15,b=7, c=5, d=10,  e=12.

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Question is incomplete as it should include figure showing table of values.

The diagram where angles 1 and 2 form a linear pair is option D which is that a horizontal line has 1 line extending from it. Angles 1 and 2 are formed.

Given that angle 1 and 2 form a linear pair.

We know that a linear pair is an angle that are formed when two lines intersect each other at a single point.

In our case when a horizontal line has 1 line extending from it and when angles 1 and 2 are formed shows a linear pair.

In first part when two lines intersect they donot form 4 angles.

In second part when 3 lines extend from a point they don't form right angle between the lines.

Hence the linear pair angles are shown by the statement that a horizontal line has 1 line extending from it. Angles 1 and 2 are formed.

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The complete table is shown below.

Complete the table as follows:

[tex]2^{2} = 2*2 = 4\\3^{2} = 3*3 = 9\\4^{2} = 4*4 = 16\\5^{2} = 5*5=25\\6^{2} =6*6=36\\7^{2} =7*7=49\\8^{2} =8*8=64\\9^{2} =9*9=81\\10^{2} =10*10=100\\11^{2} =11*11=121[/tex]

Therefore, the complete table has been shown.

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The question you are looking for is here:

Complete the table by squaring each positive x-value listed 2, 3,4,5,6,7,8,9,10,11.

Answer:

-70

Step-by-step explanation:

We are given this arithmetic sequence:

-2, -6, -10, -14, ...

And we want to find the 18th term in it.

The 18th term can be found using this formula:

1st term + common difference(desired term-1)

The desired term is the term that we are looking for. In this case, it would be the 18th term, so substitute 'desired term' with 18.

1st term + common difference(18-1)

So, let's find the first term and the common difference.

The 1st term is the first term (number) that appears in the sequence. In this case, that number would be -2.

The common difference can be found by doing second term minus first term.

Remember that we know that the first term is -2. The second term is the second number that appears in the sequence, which would be -6.

So, do -6 subtract -2.

-6 - - 2

-6 + 2

-4

The common difference is -4.

So, we can plug -2 and -4 into the formula.

-2 - 4(18-1)

Now, doing the order of operations, first, subtract 1 from 18.

-2-4(17)

Now multiply -4 and 17 together.

-2 -68

Subtract -68 from -2.

-70

The 18th term of the sequence is -70.

The probability that at least 9 customers ordered an item from the value menu is 0.927​.

In an experiment with two possible outcomes, the likelihood of precisely x successes on n repeated trials is known as the binomial probability (generally refereed as binomial experiment).

As per the given problem-

This likelihood is a binomial one, so take note. In this instance, a success is defined as a consumer placing an order from the value menu, hence we are looking for such probability between 9 to 14 success, inclusive.

The probability is indeed the complements of the likelihood of any success, ranging from 0 to 8. The procedures below can be used to figure out the probability from such a binomial distribution utilizing Excel.-

Subtract the preceding probability from 1 to get the likelihood of 9 to 14 people ordering out from value menu.

The likelihood is 1 x 0.073 = 0.927.

Therefore, the probability that at least 9 customers ordered an item from the value menu is 0.927.

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The number of weeks it will take for the budget money to run out is; 2 weeks.

The number of weeks it would take for the budget money to decrease till it runs out as required in the task can be interpreted as the x-value which corresponds to a Y-value of zero.

Hence, by setting y=0; we have;

-2x -0 = -4

-2x = -4

Divide both sides of the equation by; -2.

x = 2

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The missing information of the triangle is C = 63°, b ≈ 28.084 and c ≈ 25.084. (Correct choice: C)

In this question we have a triangle with a known side lengths and two known angle measures. First, we find the missing angle by Euclidean geometry:

C = 180° - 94° - 23°

C = 63°

Lastly, we determine the missing sides by law of sines:

[tex]b = 11 \times \frac{\sin 94^{\circ}}{\sin 23^{\circ}}[/tex]

b ≈ 28.084

[tex]c = 11 \times \frac{\sin 63^{\circ}}{\sin 23^{\circ}}[/tex]

c ≈ 25.084

The missing information of the triangle is C = 63°, b ≈ 28.084 and c ≈ 25.084. (Correct choice: C)

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[tex] \underline{ \orange{\huge \boxed{ \frak{Answer : }}}}[/tex]

Let ,

[tex] \sf \large \color{purple} y = 2 {x}^{3} - 3 {x}^{2} - 12x + 10 \: --( \: 1 \: )[/tex]

[tex] \: \: \: [/tex]

Now , Diff wrt ' x ' , we get :

[tex] \sf \: \frac{dy}{dx} = \frac{d}{dx} (2 {x}^{3} - 3 {x}^{2} - 12x + 10) \\ \sf \: \sf \: \frac{dy}{dx} = \frac{d}{dx} \: 2(3 {x}^{2} ) - \frac{d}{dx} 3 {x}^{2} - \frac{d}{dx} 12x + \frac{d}{dx} 10 \\ \sf \: \frac{dy}{dx} =2(3 {x}^{2} ) - 3(2x) - 12(1) + 0 \\ \sf \: \frac{dy}{dx} =6 {x}^{2} - 6x - 12 + 0 \\ \: \sf \red{\frac{dy}{dx} = 6 {x}^{2} 6x - 12 -- (2)}[/tex]

[tex] \: \: \: [/tex]

For maxima or minima \frac{dy}{dx} = 0

[tex] \: \: \: [/tex]

[tex] \sf \: 6 {x}^{2} - 6x - 12 = 0[/tex]

[tex] \: \: \: [/tex]

Divided by 6 on both side , we get.

[tex] \: \: \: [/tex]

[tex] \sf \: {x}^{2} - x - 2 = 0 \\ \sf \: {x}^{2} - 2x + x - 2 = 0 \\ \sf \: x(x - 2) + 1(x - 2) = 0 \\ \sf \: (x - 2)(x + 1) = 0 \\ \sf \: x - 2 = 0 \: \: \bold or \: \: x + 1 = 0 \\ \sf \fbox{x = 2 \: } \: \bold or \: \fbox{ x = - 1}[/tex]

[tex] \: \: \: [/tex]

Again Diff wrt ‘ x ’ , we get.

[tex] \sf \: \frac{d}{dx} =(\frac{dy}{dx} ) = 6\frac{d}{dx} - 6\frac{d}{dx}x - \frac{d}{dx}12 \\ \sf \: \frac{ {d}^{2}y }{ {dx}^{2} } = 6(2x) - 6(1) - 0 \\ \sf \: \sf \bold{ \frac{ {d}^{2}y }{ {dx}^{2} } =12x - 6}[/tex]

[tex] \: \: \: [/tex]

At x = 2

[tex] \: \: \: [/tex]

[tex]\sf \: \frac{ {d}^{2}y }{ {dx}^{2} } =12(2) - 6 \\ \: \: \: \sf \: = 24 - 6 \\ \: \: \: \: \sf \red{ = 18 > 0}[/tex]

At x = -1

[tex] \: \: \: [/tex]

[tex]\sf \: \frac{ {d}^{2}y }{ {dx}^{2} } =12( - 1) - 6 \\ \: \: \: \sf \: = - 12 - 6 \\ \: \: \: \: \sf \red{ = - 18 < 0 }[/tex]

[tex] \: \: \: [/tex]

x = 2 gives minima value of function.

[tex] \: \: \: [/tex]

x = -1 gives maxima value of function.

[tex] \: \: \: [/tex]

Now, put x = 2 in eqⁿ ( 1 )

[tex] \: \: \: [/tex]

[tex] \sf \: y \: minima \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2( {2})^{3} - 3 ({2})^{2} - 12(2) + 10 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: = 2(8) - 3(4) - 24 + 10 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \: \: = 16 - 12 - 24 + 10 \\\sf \: \: \: \: \: \: \: \: \: \: = - 20 + 10 \\\sf \color{red}{\boxed{ = - 10}}[/tex]

[tex] \: \: \: [/tex]

The Point of minima is ( 2 , -10 ).

[tex] \: \: \: [/tex]

Now , put x = -1 in eqⁿ ( 1 )

[tex]\sf \: y \: maxima \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2( { - 1})^{3} - 3 ({ - 1})^{2} - 12( - 1) + 10 \\\sf \color{red}{\boxed{ = 17}}[/tex]

[tex] \: \: \: [/tex]

The point of maxima value is ( -1 , 17 ).

[tex] \: \: \: [/tex]

[tex] \: \: [/tex]

Hope Helps! :)

Check the picture below.

Using translation concepts, the equation of the new line is:

[tex]y = -\frac{1}{2}x[/tex]

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

In this problem:

Hence the equation of the new line is:

[tex]y = -1 \times \frac{1}{2} \times x = -\frac{1}{2}x[/tex]

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Answer:

the answer is D- 9/k

Answer:

D- 9/k

Step-by-step explanation: