Which classification best represents a triangle with side lengths 10 in., 12 in., and 15 in.?
acute, because 102+122>152
acute, because 122+152>102
obtuse, because 102+122>152
obtuse, because 122+152>102

Answer:

23 there will be 100 paperclips leftover

Question 4

[tex]\sin 30^{\circ}=\frac{y}{14\sqrt{3}}\\\\\frac{1}{2}=\frac{y}{14\sqrt{3}}\\\\\boxed{y=7\sqrt{3}}\\\\\\\\\cos 30^{\circ}=\frac{x}{14\sqrt{3}}\\\\\frac{\sqrt{3}}{2}=\frac{x}{14\sqrt{3}}\\\\\boxed{x=21}[/tex]

Question 5

[tex]\cos 45^{\circ}=\frac{x}{28}\\\\\frac{1}{\sqrt{2}}=\frac{x}{28}\\\\x=\frac{28}{\sqrt{2}}\\\\\boxed{x=14\sqrt{2}}\\\\\\\\\sin 45^{\circ}=\frac{y}{28}\\\\\frac{1}{\sqrt{2}}=\frac{y}{28}\\\\y=\frac{28}{\sqrt{2}}\\\\\boxed{y=14\sqrt{2}}[/tex]

The line plot shown C in the figure is correct.

Given that the recorded lengths of ribbon pieces which is used in project is [tex]11\frac{1}{4},11\frac{1}{4},10\frac{3}{4},10\frac{1}{2},10\frac{7}{8},10\frac{5}{8},9\frac{7}{8},11\frac{1}{4}[/tex].

Frequency refers to the number of times an event or  value occurs. A frequency table is a table that lists items and shows how many times those items occurred.

A line plot displays the data using a no. line.

Firstly, we will fimd the frequency of each value

Values                               Frequency

9 (7/8)                                        1

10(4/8)=10(1/2)                           1

10(5/8)                                       1

10(6/8)=10(3/4)                          1

10(7/8)                                       1

11(2/8)=11(1/4)                             3

Option A is incorrect because 11(2/8) has no data values in the line plot (i.e. frequency = 0)

Option B is incorrect because 11(2/8) has no data values in the line plot (i.e. frequency = 0)

Option C is correct because the frequency distribution of the data is represented correctly in the line plot.

Option D is incorrect because 11(2/8) has 2 data values in the line plot (i.e., frequency = 2)

Hence, the line plot which correct shows the data is third line plot.

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The cost of kilograms of almonds is $8.5 if 4 kilograms of almonds cost $34.

According to the questions,

4 kilograms of almonds cost $34. In order to find one kilograms of almonds  only if we divide 34 by 4.

= [tex]\frac{34}{4}[/tex] = 8.5

Hence, the cost of kilograms of almonds is $8.5 if 4 kilograms of almonds cost $34.

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Point-slope form

[tex]y-2=4(x+1)[/tex]

Slope-intercept form

[tex]y-2=4x+4 \\ \\ y=4x+6[/tex]

Standard form

[tex]4x-y+6=0[/tex]

Answer:

[tex]\textsf{Point-slope form}: \quad \sf y-2=4(x+1)[/tex]

[tex]\textsf{Slope-intercept form}: \quad \sf y=4x+6[/tex]

[tex]\textsf{Standard form}: \quad \sf 4x-y=-6[/tex]

Step-by-step explanation:

Given information:

Point-slope form of linear equation:  

[tex]\sf y-y_1=m(x-x_1)[/tex]

(where m is the slope and (x₁, y₁) is a point on the line)

Substitute the given slope and point into the formula:

[tex]\implies \sf y-2=4(x-(-1))[/tex]

[tex]\implies \sf y-2=4(x+1)[/tex]

Slope-intercept form of a linear equation:

[tex]\sf y=mx+b[/tex]

(where m is the slope and b is the y-intercept)

Substitute the given slope and point into the formula and solve for b:

[tex]\implies \sf 2=4(-1)+b[/tex]

[tex]\implies \sf b=6[/tex]

Therefore:

[tex]\sf y=4x+6[/tex]

Standard form of a linear equation:

[tex]\sf Ax+By=C[/tex]

(where A, B and C are constants and A must be positive)

Rearrange the found slope-intercept form of the equation into standard form:

[tex]\implies \sf y=4x+6[/tex]

[tex]\implies \sf 4x-y+6=0[/tex]

[tex]\implies \sf 4x-y=-6[/tex]

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hi

"does not change "  or "remain the same "

The charges of entrance fee for a bus containing 30 people is $98

let

Charges For 2 people = mx + y

14 = 2x + y

Charges For 4 people = mx + y

20 = 4x + y

14 = 2x + y

20 = 4x + y

20 - 14 = 4x - 2x

6 = 2x

x = 6/2

x = 3

Substitute x = 3 into

14 = 2x + y

14 = 2(3) + y

14 = 6 + y

14 - 6 = y

8 = y

Charges For 8 people = mx + y

= 8(3) + 8

= 24 + 8

= $32

Therefore,

Charges For 30 people = mx + y

= 30(3) + y

= 90 + 8

= $98

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Answer:

The answer is X=3/4 or -1/2

In a two-tailed hypothesis test the null hypothesis is rejected.

According to the statement

The test statistic is 3.06 and the critical values are /-2.086.

If we find the value of t by using the formula the then it comes with Negative sign and it lies under the critical region.

That's why In a two-tailed hypothesis test the null hypothesis is rejected.

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Answer:

Since April 1, 1986 was a Tuesday, March,

1986 had? Mondays.

og

the answer A

Answer:

[tex](f-g)(x)=x^2-6x+5[/tex]  dom: (-∞,∞)

[tex](\frac{f}{g})(x)=x-7+\frac{7}{x+2}[/tex] dom: (–∞,–2)u(–2,∞)

Step-by-step explanation:

[tex](f-g)(x)=f(x)-g(x)=(x^2-5x+7)-(x+2)[/tex] *distribute the negative sign into (x+2)!

[tex]x^2-5x+7-x-2\\x^2-6x+5[/tex]

a parabola (anything that begins with [tex]x^2[/tex]) will have a domain of (-∞,∞) or all real numbers!!

[tex](\frac{f}{g} )(x)= \frac{x^2-5x+7}{x+2}[/tex]      use synthetic division to divide (the attached picture)

domain: Because the graph is not continuous, you have to write the domains on both sides of the asymptote which is (-∞,-2)u(-2,∞)

synthetic division:

1. take the divisor (x+2) and solve for x.    x= –2 this goes in the top left corner

2. write the numbers AND their signs on the top row. if there is no number and just the variable (like [tex]x^2[/tex] ) just write 1.

3. the first number gets pulled down

4. multiply -2 by 1 and subtract it from –5. (-5-2= -7)

multiply -2 by -7 and add that to the next number in the top row which is -7. (-7 + 14=7)

5. the first number in the bottom row of numbers is the first number in the answer but with one less exponent than the dividend. **write 1 as x**

6. the last number in the bottom row, if it is not 0, is a remainder. write it as that number over the divisor. in this case the remainder is 7. so write it as [tex]\frac{7}{x+2}[/tex]

No she is not correct 1/3 is great that 1/4

Answer:

No, she isn't correct because if the numerator are same but if denominators is different,then the fraction with smaller denominator is greater.

(a) The scale used in the analysis is qualitative as the data is not assigned any numeric value.

The daily performance is an ordinal variable, as the values assigned are ranked according to an order of the delivery time.

(b) The frequency table of the given data is:

Variable Frequency

Early  23

On-time  65

Late            9

Lost            3

Total 100.

(c) The relative frequency table for the given data is:

Variable Frequency Relative frequency

Early  23           23/100 = 0.23 = 23%

On-time  65           65/100 = 0.65 = 65%

Late            9             9/100 = 0.09 = 9%

Lost            3             3/100 = 0.03 = 3%

(a) The scale used in the analysis is qualitative as the data is not assigned any numeric value.

The daily performance is an ordinal variable, as the values assigned are ranked according to an order of the delivery time.

(b) The frequency table is a table showing frequency of each variable for the number of times they appear in the data.

The frequency table of the given data is:

Variable Frequency

Early  23

On-time  65

Late            9

Lost            3

Total 100.

(c) The relative frequency for a variable is the ratio of its frequency to the total frequency.

The relative frequency table for the given data is:

Variable Frequency Relative frequency

Early  23           23/100 = 0.23 = 23%

On-time  65           65/100 = 0.65 = 65%

Late            9             9/100 = 0.09 = 9%

Lost            3             3/100 = 0.03 = 3%

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Answer:

C.

Step-by-step explanation:

The expression that must be true for a to be parallel to b are mAngle2 = 58Degrees and <1 = 4x - 10

A line is the shortest distance between two points and the point where two lines meet is known as an angle

From the figure shown

m<2 = 58 degrees (alternate exterior angle)

Since the sum of angles on a straight line is 180 degrees, hence;

3x-1+<2 + 4x - 10= 180

3x-1 + 58 + 4x - 10 =180

7x + 47 = 180

7x = 180- 47

7x = 133

x = 19

Hence the expression that must be true for a to be parallel to b are mAngle2 = 58Degrees and <1 = 4x - 10

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Answer:

angle 2=58 degrees

angle 1=4x-10 degrees

Step-by-step explanation:

The equation p=4+3y is maximised at (4,5) and the maximised value is 31.

Given equation p=4x+3y and constraints x>0,x<4, y<5.

We have to maximise the equation p=4x+3y with constraints x>0,x<4,y<5.

To find out the points we have to make graphs of the constraints first.

From the graph we have found that points are (0,0),(0,5),(4,0),(4,5).

The above points are common points of graphs of all constraints.

The value of equations are as under:

At (0,0) , p=4*0+3*0=0

At (0,5), p=4*0+3*5=15

At (4,0),p=4*4+3*0=16

At (4,5),p=4*4+3*5=16+15=31

Maximum value of p is 31 and that is for (4,5).

Hence the equation is maximised at (4,5) and the maximum value is 31.

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Answer:

x=27.80

Step-by-step explanation:

we know that cosA=base/hypotaneous

so,

[tex]cos44 = \frac{20}{x} \\ x = \frac{20}{cos(44)} \\ = \frac{20}{0.7193...} \\ = 27.80 [/tex]

Question 6

1) [tex]\overline{AB} \cong \overline{BD}[/tex], [tex]\overline{CD} \perp \overline{BD}[/tex], O is the midpoint of [tex]\overline{BD}[/tex], [tex]\overline{AB} \cong \overline{CD}[/tex] (given)

2) [tex]\angle ABO, \angle ODC[/tex] are right angles (perpendicular lines form right angles)

3) [tex]\triangle ABO, \triangle CDO[/tex] are right triangles (a triangle with a right angle is a right triangle)

4) [tex]\overline{BO} \cong \overline{OD}[/tex] (a midpoint splits a segment into two congruent parts)

5) [tex]\triangle ABO \cong \triangle CDO[/tex] (LL)

Question 7

1) [tex]\angle ADC, \angle BDC[/tex] are right angles), [tex]\overline{AD} \cong \overline{BD}[/tex]

2) [tex]\overline{CD} \cong \overline{CD}[/tex] (reflexive property)

3) [tex]\triangle CDA, \triangle CDB[/tex] are right triangles (a triangle with a right angle is a right triangle)

4) [tex]\triangle ADC \cong \triangle BDC[/tex] (LL)

5) [tex]\overline{AC} \cong \overline{BC}[/tex] (CPCTC)

Question 8

1) [tex]\overline{CD} \perp \overline{AB}[/tex], point D bisects [tex]\overline{AB}[/tex] (given)

2) [tex]\angle CDA, \angle CDB[/tex] are right angles (perpendicular lines form right angles)

3) [tex]\triangle CDA, \triangle CDB[/tex] are right triangles (a triangle with a right angle is a right triangle)

4) [tex]\overline{AD} \cong \overline{DB}[/tex] (definition of a bisector)

5) [tex]\overline{CD} \cong \overline{CD}[/tex] (reflexive property)

6)  [tex]\triangle ADC \cong \triangle BDC[/tex] (LL)

7) [tex]\angle ACD \cong \angle BCD[/tex] (CPCTC)

the value of 'x' is 0. 9

For the given triangle, we have the;

We want to find the value of x

Using the tangent ratio

Tan θ = opposite/adjacent

Let's substitute the value of theta, opposite and adjacent

Tan 60 = 13 + x / 7

1. 73 = 13+ x/ 7

cross multiply

1. 73 × 7 = 13 + x

12. 1 = 13 + x

x = 13 - 12. 1

x = 0. 9

We can see that the value of x is 0. 9

Thus, the value of 'x' is 0. 9

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The probability that a prize ticket with a vowel will be chosen, not replaced, and then another prize ticket with a vowel will be chosen is 0.031 or 3.1%. So, this represents a dependent event.

It is given that there are 26 prize tickets in a bowl, labeled A to Z.

There are 5 vowels- A, E, I, O, U

So, the probability of choosing a ticket with a vowel will be - (5/26)

Another ticket with a vowel will be chosen without replacing the first ticket. Then, its probability is (4/25)

So, the required probability is,

(5/26) × (4/25) = 2/65 = 0.031

The event of choosing the first ticket with a vowel effect the event of choosing the second ticket with a vowel and vice-versa. So, this represents dependent events.

Therefore, the probability is 3.1 %.

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Answer:

33.3

Step-by-step explanation:

The scale factor is 2/3.

The ratio of the surface areas is the square of the scale factor, so the surface area of the smaller cone is 75(2/3)², which is about 33.3 cm²

Answer:

96

Step-by-step explanation:

(2^22-2^20) (3^24-3^22) (4^26-4^24)

=[(2^2-1)2^20][(3^3-1)3^22][(4^2-1)4^24]

=3*2^20*8*3^22*15*4^24

=2^23*3^23*4^24*15

=2^23*3^23*2^48*3*5

=2^71*3^24*5.

∵(2^22-2^20) (3^24-3^22) (4^26-4^24)=2^x × 3^y × 5^z，

∴x=71, y=24, z=1,

∴x+y+z=71+24+1=96.

Using the binomial distribution, it is found that there is a 0.0012 = 0.12% probability at least two of them make it inside the recycling bin.

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

With 5 shoots, the probability of making at least one is [tex]\frac{211}{243}[/tex], hence the probability of making none, P(X = 0), is [tex]\frac{232}{243}[/tex], hence:

[tex](1 - p)^5 = \frac{232}{243}[/tex]

[tex]\sqrt[5]{(1 - p)^5} = \sqrt[5]{\frac{232}{243}}[/tex]

1 - p = 0.9908

p = 0.0092

Then, with 6 shoots, the parameters are:

n = 6, p = 0.0092.

The probability that at least two of them make it inside the recycling bin is:

[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]

In which:

[P(X < 2) = P(X = 0) + P(X = 1)

Then:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{6,0}.(0.0092)^{0}.(0.9908)^{6} = 0.9461[/tex]

[tex]P(X = 1) = C_{6,1}.(0.0092)^{1}.(0.9908)^{5} = 0.0527[/tex]

Then:

P(X < 2) = P(X = 0) + P(X = 1) = 0.9461 + 0.0527 = 0.9988

[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9988 = 0.0012[/tex]

0.0012 = 0.12% probability at least two of them make it inside the recycling bin.

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Answer:

[tex]\huge\boxed{\sf x = 55\°}[/tex]

Step-by-step explanation:

Given that,

x° + 35° = 90° (complementary angles)

Subtract 35 to  both sides

x = 90 - 35

x = 55°

[tex]\rule[225]{225}{2}[/tex]

Answer:

Step-by-step explanation:

i think it is c

Answer:

A

Step-by-step explanation:

So let's look at each and determine whether they're true or not.

A. It's vertex is at the origin:

   This is true! The reason for this is because you can express a quadratic in vertex form as such: [tex]f(x)=a(x-h)+k[/tex] where (h, k) is the vertex. If you remember the parent function is simply: [tex]f(x)=x^2[/tex] which we can express as: [tex]f(x) = (x-0)+0[/tex] meaning the vertex is at (0, 0). This also intuitively makes sense, the x^2 will only output positive values for real numbers, so when x<0, f(x) is still going to be positive because it's squaring the negative, so that means when you go from -5 to -4, even though you increased x by 1, the f(x) decreases by 1, since it's squaring the value to get a positive value. The lowest value this can output is 0 because 0^2 = 0. This means you vertex is at the origin.

B. It is not a function:

   So the quadratic parent function only outputs 1 value for each input, although it isn't a one-to-one function, meaning that each output isn't unique, but each input still only outputs 1 output, which makes this a function

C. It is not a parabola:

   parabolas are expressed as quadratic equations, so this is false.

D. It is in quadrants ||| and IV:

   This is not true since the parent function opens upwards and has a vertex at (0, 0) so it will only be in quadrant I and II. I'll provide a diagram showing this.

a. The equivalent expression for cos(4x) in terms of just x is 8cos⁴x - 8cos²x + 1

b. The equivalent expression for sin(4x) in terms of just x is 4sinxcos³x - 2sinxcosx

To find the trigonometric expression for cos(4x)

cos(4x) = cos[2(2x)]

Now from trigonometric identities

cos2Ф = 2cos²Ф - 1

Let 2x = Ф.

So, cos(4x) = 2cos²2x - 1

Also, cos2x = 2cos²x - 1

So, substituting cos2x into cos4x, we have

cos(4x) = 2cos²2x - 1

cos(4x) = 2[2cos²x - 1]² - 1

cos(4x) = 2[4cos⁴x - 4cos²x + 1] - 1

cos(4x) = 8cos⁴x - 8cos²x + 2 - 1

cos(4x) = 8cos⁴x - 8cos²x + 1

So, the equivalent expression for cos(4x) in terms of just x is 8cos⁴x - 8cos²x + 1

To find the expression for sin(4x)

sin(4x) = sin2(2x)

From trigonometric identities sin2Ф = 2sinФcosФ

Let 2x = Ф

So, sin(4x) = sin2(2x)

= 2sin2xcos2x

Since

Substituting these into the equation, we have

sin(4x) =  2sin2xcos2x

sin(4x) = (2sinxcosx)(2cos²x - 1)

sin(4x) = 4sinxcos³x - 2sinxcosx

So, the equivalent expression for sin(4x) in terms of just x is 4sinxcos³x - 2sinxcosx

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The length of the side of the octagon is 4.142 inches.

Let the length of the congruent sides of the isosceles triangle be x.

The hypotenuse of the triangle is given by:

√x+x = √2x

This is also the length of the sides of the octagon. It is given that the sides of the octagon are equal.

It is also given that the length of the square side is 10 inches.

This means that:

x + √2x + x = 10

2x + √2x = 10

(2 + √2)x = 10

(2 + 1.414)x = 10

3.414x = 10

x = 10/3.414

x = 2.929 inches.

The side of the octagon is given by:

√2 × 2.929 = 4.142 inches

Therefore, we have found that the length of the sides of the octagon is 4.142 inches.

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Answer:

correct answer is 110x + 29y = 333

Step-by-step explanation:

Step-by-step explanation:

we first divide fourteen by four

14/4= 3.5

then we change 0.5 to hours by multiplying by 60

0.5 * 60 = 30

so in conclusion it would take 3hrs 30mins to finish each activity

Answer:

They are similar.  Triangle abc is similar to triangle edc.

Step-by-step explanation:

To prove that triangles are similar, you must show that the 3 angles in one triangle are equal to the 3 angles in the other triangle.  We know that they both have a 50 degree angle.  We can also see that they both have a right angle.  That means that the 3rd angle must also equal 40 (180-90-50).  So by AAA the two triangles are congruent.

Triangle ABC is similar to triangle EDC.

Answer:

triangle ABC is similar to triangle DCE

Step-by-step explanation:

Reason being that angles in triangle ABC Correspond to That of triangle DCE Eg triangle ABC=CDE