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2 years ago

Given h(x) = 5x + 2, solve for 2 when h(x) = -8.

Mathematics

1 answer:

san4es73 [151]2 years ago

8 0

Answer:

$\huge\boxed{\sf x = -2}$

Step-by-step explanation:

$\sf h(x) = 5x+2\\\\Put \ h(x) = -8\\\\-8 = 5x+2\\\\Subtract \ 2 \ to \ both \ sides\\\\-8-2 = 5x\\\\-10 = 5x\\\\Divide\ both \ sides \ by \ 5\\\\-10 / 5 = x \\\\x = -2 \\\\\rule[225]{225}{2}$

Hope this helped!

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3 years ago

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A survey of 500 high school students was taken to determine their favorite chocolate candy. Of the 500 students surveyed, 150 li

Anika [276]

Answer:

N(AUC∩B') = 121

The number of students that like Reese's Peanut Butter Cups or Snickers, but not Twix is 121

Step-by-step explanation:

Let A represent snickers, B represent Twix and C represent Reese's Peanut Butter Cups.

Given;

N(A) = 150

N(B) = 204

N(C) = 206

N(A∩B) = 75

N(A∩C) = 100

N(B∩C) = 98

N(A∩B∩C) = 38

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How many students like Reese's Peanut Butter Cups or Snickers, but not Twix;

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This can be derived by first finding;

N(AUC) = N(A) + N(C) - N(A∩C)

N(AUC) = 150+206-100 = 256

Also,

N(A∩B U B∩C) = N(A∩B) + N(B∩C) - N(A∩B∩C) = 75 + 98 - 38 = 135

N(AUC∩B') = N(AUC) - N(A∩B U B∩C) = 256-135 = 121

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The number of students that like Reese's Peanut Butter Cups or Snickers, but not Twix is 121

See attached venn diagram for clarity.

The number of students that like Reese's Peanut Butter Cups or Snickers, but not Twix is the shaded part

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2 years ago

Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A

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Answer:

$\theta_{CAB}=128.316$

$\theta_{ABC}=25.842$

$\theta_{BCA}=25.842$

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

$r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

$AB= \sqrt{(-3-1)^2+(0-3)^2}=5$

$BC= \sqrt{(1-1)^2+(3-(-3))^2}=9$

$CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5$

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

$BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}$

$\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}$

$\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}$

$\theta_{CAB}=\arccos{-\dfrac{0.62}}$

$\theta_{CAB}=128.316$

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

$\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}$

$\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)$

$\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)$

$\theta_{ABC}=\arcsin{0.4359}\right)$

$\theta_{ABC}=25.842$

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

$\theta_{BCA}=25.842$

this can also be checked using the fact the sum of all angles inside a triangle is 180

$\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180$

$25.842+128.316+25.842$

$180$

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Answer:

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The derivative of R with respect to r1 is:

$\frac{dR}{dr_1}=\frac{r_2(r_1+r_2)-1(r_1r_2)}{(r_1+r_2)^2}=\frac{r_2^2}{(r_1+r_2)^2}$

We notice that the derivative is a fraction of two squared terms: therefore, both factors are positive, so the derivative is always positive, and this means that R is an increasing function of r1.

To solve this part, we use again the expression for R written in part a:

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We start by noticing that there is a limit on the allowed values for r1: in fact, r1 must be strictly positive,

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