In ΔABC, BD is perpendicular to AC as shown in the figure.

Which equations are true?

(C) [tex]\frac{1}{3}[/tex] fraction of the marbles now in the bag is blue.

To find what fraction of the marbles now in the bag is blue:

Therefore, (C) [tex]\frac{1}{3}[/tex] fraction of the marbles now in the bag is blue.

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Complete question:

A bag initially contains red marbles and blue marbles only, with more blue than red. Red marbles are added to the bag until only 1 / 3 of the marbles in the bag are blue. Then yellow marbles are added to the bag until only 1 / 5 of the marbles in the bag are blue. Finally, the number of blue marbles in the bag is doubled. What fraction of the marbles now in the bag are blue?

(A) 1/5

(B) 1/4

(C) 1/3

(D) 2/5

(E) 1/2

When two or more given triangles are congruent, this implies that the measures of their corresponding angles and sides are equal. So that the additional information required in the question is FG ≅ IJ.

When two or more given triangles are congruent, this implies that the measures of their corresponding angles and sides are equal. Thus the congruent relations of the triangles can be with respect to their angles or/ and sides.

The two given triangles in the question would be congruent with respect to the Side-Angle-Side (SAS) postulate if the condition below is true:

i.e FG ≅ IJ

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Answer:

[tex]x = \frac{b}{6a} - c[/tex]

Step-by-step explanation:

Steps are as below:

(I will refer A as 'a' and B as 'b')

[tex]a = \frac{b}{c + x} - 5a \\ a + 5a = \frac{b}{c + x} \\ 6a = \frac{b}{c + x} \\ 6a(c + x) = b \\ 6ac + 6ax = b \\ 6ac = b - 6ax \\ b - 6ax = 6ac \\ - 6ax = 6ac - b \\ 6ax = - (6ac - b) \\ 6ax = b - 6ac \\ x = \frac{b - 6ac}{6a} \\ x = \frac{b}{6a} - \frac{6ac}{6a} \\ x = \frac{b}{6a} - c[/tex]

The rate at which the water from the container is being drained is 24 inches per second.

Given radius of right circular cone 4 inches .height being 5 inches, height of water is 2 inches and rate at which surface area is falling is 2 inches per second.

Looking at the image we can use similar triangle propert to derive the relationship:

r/R=h/H

where dh/dt=2.

Thus r/5=2/5

r=2  inches

Now from r/R=h/H

we have to  write with initial values of cone and differentiate:

r/5=h/5

5r=5h

differentiating with respect to t

5 dr/dt=5 dh/dt

dh/dt is given as 2

5 dr/dt=5*-2

dr/dt=-2

Volume of cone is 1/3 π[tex]r^{2} h[/tex]

We can find the rate at which the water is to be drained by using partial differentiation on the volume equation.

Thus

dv/dt=1/3 π(2rh*dr/dt)+([tex]r^{2}[/tex]*dh/dt)

Putting the values which are given and calculated we get

dv/dt=1/3π(2*2*2*2)+(4*2)

=1/3*3.14*(16+8)

=3.14*24/3.14

=24 inches per second

Hence the rate at which the water is drained from the container is 24 inches per second.

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Answer:

base = 12 cm , altitude = 16 cm

Step-by-step explanation:

the area (A) of a triangle is calculated as

A = [tex]\frac{1}{2}[/tex] ba ( b is the base and a the altitude )

here a = b + 4 , then

[tex]\frac{1}{2}[/tex] b(b + 4) = 96 ( multiply both sides by 2 to clear the fraction )

b(b + 4) = 192

b² + 4b = 192 ( subtract 192 from both sides )

b² + 4b - 192 = 0 ← in standard quadratic form

(b + 16)(b - 12) = 0 ← in factored form

equate each factor to zero and solve for b

b + 16 = 0 ⇒ b = - 16

b - 12 = 0 ⇒ b = 12

however, b > 0 then b = 12

so base = 12 cm and altitude = b + 4 = 12 + 4 = 16 cm

Answer:

base = 12 cm

Altitude = 16 cm

Step-by-step explanation:

        [tex]\sf \boxed{Area \ of \ triangle = \dfrac{1}{2}*base*height}[/tex]

base = x cm

altitude or height = (x + 4) cm

[tex]\sf \dfrac{1}{2}*x*(x+4) = 96\\\\[/tex]

x * (x + 4) = 96*2

x*x + x*4 = 192

x² + 4x   = 192

x² + 4x - 192 = 0

Sum = 4

Product = -192

Factors =  (-12) , 16

When we add (-12) & 16, we get 4. When we multiply (-12) & 16, we get (-192).

Rewrite the middle term using the factors.

x² + 16x - 12x - 192 = 0

x(x + 16) -12(x + 16) = 0

(x + 16)(x - 12) = 0

x - 12 = 0     or x + 16 = 0

x = 12             or x = -16

Ignore x = -16 as measurements won't be in negative value.

x = 12

Base = 12 cm

Altitude = 12 + 4 = 16 cm

Answer: G=4

Step-by-step explanation:

Answer:

I and II only

Answer:

∠α = 5.001°

a easy calculator for this is:

https://www.calculator.net/right-triangle-calculator.html?av=3.5&alphav=&alphaunit=d&bv=40&betav=&betaunit=d&cv=&hv=&areav=&perimeterv=&x=57&y=16

Q1

Nominal - D

Ordinal-  B, G

Ratio - A, C, E, F

Q2

Quantitative- A, C, D, G

Qualitative -B, E, F, G

Q3

Simple Random - A, C

Systematic Sample - B

Nominal measures are the most accurate and meaningless, as they only show the "characteristics" or "identities" that we are interested in. In other words, in a nominal variable, a number just gives the attribute a unique "name". In this case, the number is just a label.

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The inequality passing through (1, 4), (-2, -2), intersecting the y-axis at 2 units and intersects the x-axis at (-1, 0) to form shaded portions on the left side is y > 2x+2

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

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

The inequality passing through (1, 4), (-2, -2), intersecting the y-axis at 2 units and intersects the x-axis at (-1, 0) to form shaded portions on the left side is y > 2x+2

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The wide-area search method employs the use of markers such as knots at regular intervals along the search line.

Given the method employs the use of markers such as knots at regular intervals along the search line to indicate distance from the beginning of the search line.

In order to locate, relieve distress, and preserve the life of a person who has been reported missing or is believed to be lost, stranded, or is considered a high-risk missing person, wide area search and rescue refers to activities occurring within large geographic areas. It also refers to the removal of any survivors to a safe location.

Hence, the wide-area search method employs the use of markers such as knots at regular intervals along the search line to indicate distance from the beginning of the search line

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The length of the square pool is 8. 49cm

Note that the formula for finding the area of a square is given as;

Area = length^2

We have

Area = 72 m²

Substitute into the formula

I^2 = 72

Find the square root of both sides

I= √72

I = 8. 49cm

The length of the square pool given as I is 8. 49cm

Thus, the length of the square pool is 8. 49cm

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The blue marble is predicted to be picked 120 times, in the experiment of picking a marble from a bag containing 3 red and 2 blue marbles and performing this experiment 300 times, using the theoretical probability.

The theoretical probability of any event is the ratio of the number of favorable outcomes to the event, to the total number of possible outcomes in the experiment.

If we have an event A, the number of favorable outcomes to event A as n, and the total number of possible outcomes in the experiment as S, then the theoretical probability of event A is given as:

P(A) = n/S.

In the question, we are given that Ellen has a bag with 3 red marbles and 2 blue marbles in it. She is going to randomly draw a marble from the bag 300 times, putting the marble back in the bag after each draw.

We are asked the predict the number of times that the marble picked will be blue using the theoretical probability.

Let the event of picking a blue marble be A.

The number of favorable outcomes to event A (n) = 2 {The total number of blue marbles in the bag}.

The total number of possible outcomes in the experiment of picking a ball (S) = 5 {The total number of marbles in the bag}.

Thus, the theoretical probability of event A is,

P(A) = n/S = 2/5 = 0.4.

To predict the number of times marble picked was blue, we multiply the time's the experiment was performed by the theoretical probability of picking a blue ball.

Thus, the predicted number of times = 300 * 0.4 = 120.

Thus, the blue marble is predicted to be picked 120 times, in the experiment of picking a marble from a bag containing 3 red and 2 blue marbles and performing this experiment 300 times, using the theoretical probability.

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1) JKLM is a parallelogram, [tex]\overline{KL} \cong \overline{LM}[/tex] (given)

2) JKLM is a rhombus (a parallelogram with a pair of consecutive congruent sides is a rhombus)

Based on the angle of the seesaw and the height off the ground, the maximum length of the seesaw is 2 meters.

This can be found as:

Sin 30° = 1 / Maximum length

Solving gives:

Sin 30° = 1 / Maximum length

Ml = 1 / Sin 30°

= 1 / 0.5

= 2 meters

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The researchers could utilize a probability sample.

Meaning of a Probability Sample

Every member of a population must have an equal chance of being chosen for a study in order for a sample to be considered a probability sample, and the researcher must be aware of this likelihood. The most typical type of sampling used in public opinion research, election polling, and other studies whose findings will be generalized to a larger population is probability sampling in which probability samples are made use of.

Probability Sample as the Representative of Population Sample

Here, a sample that qualifies for a representative of the population the researchers are studying, must consist of members having equal chances to be chosen for the study. Thus, a probability sample is the best representative of their study population.

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Answer:

The longest side of room is 24m

Answer:

? = 38°

Step-by-step explanation:

We can use the property of sine to find the missing angle value.

sin (angle) = opposite / hypotenuse

sin (?) = 16 / 26                                  <----- Insert values

? = sin⁻¹(16 / 26)                                 <----- Rearrange

? = 37.97...                                         <----- Solve

? = 38°                                               <----- Round

The derivative of the natural logarithm function is given by:

[tex]f^{\prime\prime}(x) = \cotg{x}[/tex]

Suppose we have a composite natural logarithm function given as follows:

[tex]f(x) = \ln{g(x)}[/tex]

The derivative of the function is given as follows:

[tex]f^{\prime}(x) = \frac{g^{\prime}(x)}{g(x)}[/tex]

In this problem, the function, which is already a first derivative, it given by:

[tex]f^{\prime}(x) = \ln{|\sin{x}|} + 3[/tex]

The derivative of a constant is of zero, and the derivative of sin(x) is cos(x), hence the derivative of the entire function is given as follows:

[tex]f^{\prime\prime}(x) = \frac{\cos{x}}{\sin{x}} = \cotg{x}[/tex]

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The number of degrees that the direction of the boat change when it made its first turn is 39°.

From the information given, we would use the cosine function to solve the information.

This will be:

18² = 28² + 25² - 2(28)(25)(cos x)

324 = 784 + 625 - 1400cosx

x = 39.19

Therefore, the value is 39°.

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Answer:

Step-by-step explanation:

70 = x/4

x = 4 cos 70 = 1.37m

x = 4 cos  80 = 0.69m  

The range is ( 0.69m, 1.37m)

The range of distances from the wall (x) that the foot of the ladder  placed to fall within the safety guidelines is between 0.7052 meters and 1.456 meters.

To determine the range of distances from the wall that the foot of the ladder may be placed to fall within the safety guidelines,  use trigonometry.

Let's assume the distance from the wall to the foot of the ladder is x meters the ladder is 4 meters long.

The angle between the ladder and the ground is given to be between 70° and 80°. Let's consider the extreme cases for each angle:

When the ladder makes an angle of 70° with the ground:

In this case, the angle between the wall and the ladder (θ) will be 90° - 70° = 20°.

When the ladder makes an angle of 80° with the ground:

In this case, the angle between the wall and the ladder (θ) will be 90° - 80° = 10°.

Now, use trigonometry to calculate the range of distances (x) from the wall:

For the first case (θ = 20°):

tan(20°) = Opposite / Adjacent

tan(20°) = x / 4

x = 4 × tan(20°)

For the second case (θ = 10°):

tan(10°) = Opposite / Adjacent

tan(10°) = x / 4

x = 4 × tan(10°)

Now, let's calculate the values:

x ≈ 4 × 0.3640 ≈ 1.456 meters (rounded to three decimal places) - for the first case.

x ≈ 4 × 0.1763 ≈ 0.7052 meters (rounded to four decimal places) - for the second case.

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The perimeter of the polygon that circumscribes the circle is 78 cm

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

The circle tangent theorem states that tangents which meet at the same point are equal in length. Hence:

The missing lengths are 8, 6, 9 and 16.

Perimeter of the polygon = 2(8) + 2(6) + 2(9) + 2(16) = 78 cm

The perimeter of the polygon that circumscribes the circle is 78 cm

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There can be 18, 12, 9, 6, 4 , 3 or 2 students on each team depending on the number of teams selected.

Considering there are at least 2 teams but no more than 19 teams, let's calculate the number of students in each case:

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The relationship that represents the law of cosines if the measure of the included angle between the sides a and b is D. area of square c² = area of square a² area of squareb² area of defect1 area of defect2

It should be noted that the cosine law is simply used to solving triangles.

The law simply states that c² = a² + b² - 2abcosC.

The correct option that illustrates this is D.

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Answer:

c. area of square c2 = area of square a2 + area of square b2 – area of defect1 – area of defect2

Step-by-step explanation:

There is no change in the temperature.

To find how much temperature is changed:

Let, 'F' be the temperature.

Temperature decreased by [tex]33\frac{1}{3}[/tex]%.

F × (1-.33333) = Temperature increases by 50%.

(F × (1-.33333)(1.50)) = F =  No net change, right back to the original temperature F.

Therefore, there is no change in the temperature.

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Answer:

Step-by-step explanation:

$1,248.92