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2 years ago

1. What is the slope of a line that is parallel to the line represented by the equation y=-2/5x + 4/5

Mathematics

1 answer:

lbvjy [14]2 years ago

4 0

Hi student, let me help you out! :)

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We are asked to find the slope of the line that is parallel to the line $\star~\mathrm{y=-\cfrac{2}{5}x+\cfrac{4}{5}}$ .

$\triangle~\fbox{\bf{KEY:}}$

Parallel lines have the same slope.

So let's first determine the slope of the first line, and then let's find the slope of the line that's parallel to it.

$\circ\!\!\!\star\mathcal{HINT:}$

The slope is the coefficient of x.

So the slope is

$\star~\mathrm{-\cfrac{2}{5}}$ .

Can you guess the slope of the line that is parallel to the first line?

That's right! The slope is...

$\star~\mathrm{-\cfrac{2}{5}}$

Hope it helps you out! :D

Ask in comments if any queries arise.

#StudyWithBrainly

~Just a smiley person helping fellow students :)

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Marina86 [1]

Answer:H

Step-by-step explanation:

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3 years ago

which of the functions have a range of real numbers greater than or equal to 1 or less than or equal to-1

lilavasa [31]

The question is incomplete. Here is the complete question:

Which of the functions have a range of all real numbers greater than or equal to 1 or less than or equal to -1? check all that apply.

A. $y=\sec x$

B. $y= \tan x$

C. $y= \cot x$

D. $y= \csc x$

Answer:

A. $y=\sec x$

D. $y=\csc x$

Step-by-step explanation:

Given:

The range is greater than or equal to 1 or less than or equal to -1.

The given choices are:

Choice A: $y=\sec x$

We know that, the $\sec x=\frac{1}{\cos x}$

The range of $\cos x$ is from -1 to 1 given as [-1, 1]. So,

$|\cos x|\leq 1\\\textrm{Taking reciprocal, the inequality sign changes}\\\frac{1}{|\cos x|}\geq 1\\|\sec x|\geq 1$

Therefore, on removing the absolute sign, we rewrite the secant function as:

$\sec x\leq -1\ or\ \sec x\geq 1\\$

Therefore, the range of $y=\sec x$ is all real numbers greater than or equal to 1 or less than or equal to-1.

Choice B: $y= \tan x$

We know that, the range of tangent function is all real numbers. So, choice B is incorrect.

Choice C: $y= \cot x$

We know that, the range of cotangent function is all real numbers. So, choice C is incorrect.

Choice D: $y=\csc x$

We know that, the $\csc x=\frac{1}{\sin x}$

The range of $\sin x$ is from -1 to 1 given as [-1, 1]. So,

$|\sin x|\leq 1\\\textrm{Taking reciprocal, the inequality sign changes}\\\frac{1}{|\sin x|}\geq 1\\|\csc x|\geq 1$

Therefore, on removing the absolute sign, we rewrite the cosecant function as:

$\csc x\leq -1\ or\ \csc x\geq 1\\$

Therefore, the range of $y=\csc x$ is all real numbers greater than or equal to 1 or less than or equal to-1.

6 0

3 years ago

Find a nonzero vector orthogonal to the plane through the points: ????=(0,0,1), ????=(−2,3,4), ????=(−2,2,0).

ser-zykov [4K]

Answer:

The nonzero vector orthogonal to the plane is <-9,-8,2>.

Step-by-step explanation:

Consider the given points are P=(0,0,1), Q=(−2,3,4), R=(−2,2,0).

$\overrightarrow {PQ}==$

$\overrightarrow {PR}==$

The nonzero vector orthogonal to the plane through the points P,Q, and R is

$\overrightarrow n=\overrightarrow {PQ}\times \overrightarrow {PR}$

$\overrightarrow n=\det \begin{pmatrix}i&j&k\\ \:\:\:\:\:-2&3&3\\ \:\:\:\:\:-2&2&-1\end{pmatrix}$

Expand along row 1.

$\overrightarrow n=i\det \begin{pmatrix}3&3\\ 2&-1\end{pmatrix}-j\det \begin{pmatrix}-2&3\\ -2&-1\end{pmatrix}+k\det \begin{pmatrix}-2&3\\ -2&2\end{pmatrix}$

$\overrightarrow n=i(-9)-j(8)+k(2)$

$\overrightarrow n=-9i-8j+2k$

$\overrightarrow n=$

Therefore, the nonzero vector orthogonal to the plane is <-9,-8,2>.

8 0

3 years ago

The school band is ordering health bars to sell for a fundraiser. The company that sells the bars charges $0.40 per bar plus $20

Pepsi [2]

Answer: They can order from 0 to 449 bars, but no more.

Step-by-step explanation:

Given: The school band is ordering health bars to sell for a fundraiser. The company that sells the bars charges $0.40 per bar plus $20.00 for shipping regardless of the size of the order. The band must spend less than $200.00 on bars for the fundraiser.

The inequality below relates x, the number of bars that could be ordered, with the shipping costs and their spending requirements.

$0.40x+20$