All categories

3 years ago

What is an equation for the line parallel to y = -3x – 5 that contains (-1, 8)?

Mathematics

2 answers:

jeka57 [31]3 years ago

5 0

<h3>Answer: y = -3x+5</h3>

slope = -3

y intercept = 5

==================================================

Explanation:

The given line is in the form y = mx+b, so the slope is m = -3

Any parallel line to this one will also have the same slope, but a different y intercept.

Plug (x,y) = (-1,8) and m = -3 into y = mx+b so we can find new y intercept.

---------------------

y = mx+b

8 = -3(-1) + b

8 = 3+b

8-3 = b

5 = b

b = 5

So because m = -3 and b = 5, this means we have y = -3x+5 as our answer.

grin007 [14]3 years ago

3 0

Answer: y = -3x + 5

Explanation:

For a line to be parallel the slope needs to be the same.

And we are given a point to work with:

So let’s use point-slope form:

We know that the slope is -3
And the point is (-1,8)

And the form:
y - y1 = m(x - x1)
y - 8 = -3(x -(-1))
y - 8 = -3(x + 1)
y - 8 = -3x - 3
y = -3x - 3 + 8
y = -3x + 5

Therefore, our equation is y = -3x + 5

You might be interested in

Help fast please....

Nataly [62]

Answer:

46 yd²

Step-by-step explanation:

5 0

3 years ago

Consider a value to be significantly low if its z score less than or equal to minus−2 or consider a value to be significantly hi

katrin2010 [14]

Answer:

Test scores of 10.2 or lower are significantly low.

Test scores of 31.4 or higher are significantly high.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 20.8, \sigma = 5.3$

Identify the test scores that are significantly low or significantly high.

Significantly low

Z = -2 and lower.

So the significantly low scores are thoses values that are lower or equal than X when Z = -2. So

$Z = \frac{X - \mu}{\sigma}$

$-2 = \frac{X - 20.8}{5.3}$

$X - 20.8 = -2*5.3$

$X = 10.2$

Test scores of 10.2 or lower are significantly low.

Significantly high

Z = 2 and higher.

So the significantly high scores are thoses values that are higherr or equal than X when Z = 2. So

$Z = \frac{X - \mu}{\sigma}$

$2 = \frac{X - 20.8}{5.3}$

$X - 20.8 = 2*5.3$

$X = 31.4$

Test scores of 31.4 or higher are significantly high.

3 0

3 years ago

Pleaseee helppppppppppppppppppp

quester [9]

To find Angle A we use cosine

cos ∅ = adjacent / hypotenuse

From the question

The adjacent is 17

The hypotenuse is 38

So we have

cos A = 17/38

A = cos-¹ 17/38

A = 63.4

<h3>A = 63° to the nearest degree</h3>

To find Angle C we use sine

sin ∅ = opposite / hypotenuse

From the question

The opposite is 17

The hypotenuse is 38

So we have

sin C = 17/38

C = sin-¹ 17/38

C = 26.57

<h3>C = 27° to the nearest degree</h3>

Hope this helps you

7 0

3 years ago

Which set of numbers could represent the lengths of the sides of a right triangle? 16, 32, 36 6, 7, 8 5, 12, 13 8, 12, 16

Olenka [21]

Right triangles satisfy Pythagorean equation given:
c^2=a^2+b^2
where
c is the hypotenuse
a and b are the legs
thus, from the choices given, the Pythagorean triple will be:
c] 5^2+12^2=13^2

8 0

3 years ago

Find the laplace transform of f(t) = cosh kt = (e kt + e −kt)/2

iren2701 [21]

Hello there, hope I can help!

I assume you mean $L\left\{\frac{ekt+e-kt}{2}\right\}$
With that, let's begin

$\frac{ekt+e-kt}{2}=\frac{ekt}{2}+\frac{e}{2}-\frac{kt}{2} \ \textgreater \ L\left\{\frac{ekt}{2}-\frac{kt}{2}+\frac{e}{2}\right\}$

$\mathrm{Use\:the\:linearity\:property\:of\:Laplace\:Transform}$
$\mathrm{For\:functions\:}f\left(t\right),\:g\left(t\right)\mathrm{\:and\:constants\:}a,\:b$
$L\left\{a\cdot f\left(t\right)+b\cdot g\left(t\right)\right\}=a\cdot L\left\{f\left(t\right)\right\}+b\cdot L\left\{g\left(t\right)\right\}$
$\frac{ek}{2}L\left\{t\right\}+L\left\{\frac{e}{2}\right\}-\frac{k}{2}L\left\{t\right\}$

$L\left\{t\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{t\right\}=\frac{1}{s^2} \ \textgreater \ L\left\{t\right\}=\frac{1}{s^2}$

$L\left\{\frac{e}{2}\right\} \ \textgreater \ \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{a\right\}=\frac{a}{s} \ \textgreater \ L\left\{\frac{e}{2}\right\}=\frac{\frac{e}{2}}{s} \ \textgreater \ \frac{e}{2s}$

$\frac{ek}{2}\cdot \frac{1}{s^2}+\frac{e}{2s}-\frac{k}{2}\cdot \frac{1}{s^2}$

$\frac{ek}{2}\cdot \frac{1}{s^2} \ \textgreater \ \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \ \frac{ek\cdot \:1}{2s^2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{ek}{2s^2}$

$\frac{k}{2}\cdot \frac{1}{s^2} \ \textgreater \ \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \ \frac{k\cdot \:1}{2s^2} \ \textgreater \ \mathrm{Apply\:rule}\:1\cdot \:a=a$
$\frac{k}{2s^2}$

$\frac{ek}{2s^2}+\frac{e}{2s}-\frac{k}{2s^2}$

Hope this helps!

3 0

4 years ago