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2 years ago

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7. Write the equation of the line that is perpendicular to y = -5x + 1 and passes through the point (2,-1). slope intercept equa

tion

1 answer:

Svetach [21]2 years ago

4 0

Answer:

y = 1/5x - 7/5

Step-by-step explanation:

y = 1/5x + b

-1 = 1/5(2) + b

-1 = 2/5 + b

-7/5 = b

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What is the slope of the line that passes through points (2,1) and (17, -17) write your answer in simplest form

andriy [413]

Answer:m=-6/5

Step-by-step explanation: To figure this out you need to use the slope formula.

m=y2-y1/x2-x1

m=-17-1/17-2

m=-18/15

This Can then be reduced to m=-6/5

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3 years ago

A man has mislaid his wallet. He thinks there is a 0.4 chance that the wallet is somewhere in his bedroom, a 0.1 chance it is in

lana [24]

Answer:

a. Probability = 0.15

b. Probability = 0.3

Step-by-step explanation:

Given

$P(Bedroom) = 0.4$

$P(Kitchen) = 0.1$

$P(Bathroom) = 0.2$

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Solving (a): Probability of being somewhere else

This is calculated by subtracting the sum of given probabilities from 1.

$Probability = 1 - (0.4 + 0.1 + 0.2 + 0.15)$

$Probability = 1 - 0.85$

$Probability = 0.15$

Solving (b): Probability of being in bedroom or kitchen

This is calculated as:

$Probability = P(Bedroom) + P(Kitchen)$

$Probability = 0.2 + 0.1$

$Probability = 0.3$

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2 years ago

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2 years ago

Read 2 more answers

When X=18, Y=14. Find the value of x when y=42.<br><br> A) 32<br> B) 18<br> C) -18 <br> D) 54

Karolina [17]

Answer:

D)54

Explanation:

hope this helps you...

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2 years ago

For what values of x does the curve y^2x^3-15x^2=4 have horizontal tangent lines

enyata [817]

We have the following curve:<span>

$y^2x^3-15x^2=4$

To find the tangent lines to the curve we need to use the concept of derivative, but we can’t solve this problem for $y$ , thus, let's apply implicit differentiation, so:

$\frac{d}{dx}(y^2x^3-15x^2=4) \\ \\
\frac{d}{dx}(y^2x^3)-\frac{d}{dx}(15x^2)=\frac{d}{dx}(4) \\ \\
(y^{2})'x^3+(x^3)'y^2-15(x^2)'=0 \\ \\ 2yy'x^3+3x^2y^2-30x=0 \\ \\
y'=\frac{dy}{dx}=\frac{30x-3x^2y^2}{2x^3y}=\frac{x(30-3xy^2)}{2x^3y} \\ \\
y'=\frac{30-3xy^2}{2x^2y}$

Therefore the horizontal lines occurs when $y'=0$ , then:

$\frac{30-3xy^2}{2x^2y}=0 \\ \\ That \ is, \
when: \\ \\ 30-3xy^2=0 \\ \\ \therefore xy^2=10 \rightarrow y^2=\frac{10}{x}$

If we substitute this in the original equation we have:

$\frac{10}{x}x^3-15x^2=4 \\ \\ \therefore
10x^2-15x^2=4 \\ \\ \therefore x^2=-\frac{4}{5}$

This is an absurd result because it is impossible for a squared number to get a negative number. So the conclusion is that there is no any value of $x$ in which the curve has horizontal tangent lines.</span>

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2 years ago