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2 years ago

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⦁ Write an equation of a line in slope intercept form (y = mx + b) with slope 1/3 going through the point (-6, 2). Show your wor

k for full credit.

1 answer:

Zepler [3.9K]2 years ago

3 0

$(\stackrel{x_1}{-6}~,~\stackrel{y_1}{2})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{\cfrac{1}{3}}(x-\stackrel{x_1}{(-6)}) \\\\\\ y-2=\cfrac{1}{3}(x+6)\implies y-2=\cfrac{1}{3}x+2\implies y=\cfrac{1}{3}x+4$

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Answer:

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Area: 12.56 yd²

Step-by-step explanation:

The circumference is given by ...

C = πd

C = (3.14)(4 yd) = 12.56 yd

The area is given in terms of diameter by ...

A = (π/4)d²

A = (3.14/4)(4 yd)² = 12.56 yd²

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3 years ago

Find an expression which represents the sum of (-8x+7y)(−8x+7y) and (2x-2y)(2x−2y) in simplest terms.

Yakvenalex [24]

Answer:

(-8x+7y)(−8x+7y)+(2x-2y)(2x−2y)= simplified - 68x^2+53y^2−120xy

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Step-by-step explanation:

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3 years ago

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jekas [21]

Answer:

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Step-by-step explanation:

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3 years ago

At present the sum of Monica's age and daughter's age is 44 years. . After 2 years Monica's age will be three times that if her

sergeinik [125]

Answer:

Monica's age = 34 years

Her daughter's age = 10 years

Step-by-step explanation:

Let Monica's age = x

Let her daughter'sage = y

x + y = 44 -------> equ 1

After 2 years:

Monica's age = x + 2

Her daughter's age = y + 2

x + 2 = 3 * (y+2)

x + 2 = 3y + 6

x = 3y + 6 -2

x = 3y + 4

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3y + 4 + y = 44

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x = 44 -10

x = 34

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3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%5Clarge%7B%5Cbold%20%5Cred%7B%20%5Csum%20%5Climits_%7B8%7D%5E%7B4%7D%20%7Bx%7D%5E%7B2%7D%2

kiruha [24]

Answer:

No solution

Step-by-step explanation:

We have

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

For the sum it is not correct to assume

$\sum_{x=8}^{4}x^2= 8^2 + 7^2+6^2+5^2+4^2 = 64+49+36+25+16 = 190$

Note that for

$\sum_{x=a}^b f(x)$

it is assumed $a\leq x \leq b$ and in your case $\nexists x\in\mathbb{Z}: a\leq x\leq b$ for $a>b$

In fact, considering a set $S$ we have

$\sum_{x=a}^b (S \cup \varnothing) = \sum_{x=a}^b S + \sum_{x=a}^b \varnothing$ that satisfy $S = S \cup \varnothing$

This means that, by definition $\sum_{x=a}^b \varnothing = 0$

Therefore,

$\sum_{x=8}^{4}x^2 = 0$

because the sum is empty.

For

$9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

we have other problems. Actually, this case is really bad.

Note that $\cos^2(\infty)$ has no value. In fact, if we consider for the case

$\lim_{x \to \infty} \cos^2(x)$ , the cosine function oscillates between $[-1, 1]$ , and therefore it is undefined. Thus, we cannot evaluate

$9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

and then

$\sum_{x=8}^{4}x^2 + 9\left(\dfrac{\cos^2(\infty)}{\sin^2(\infty)} \right)$

has no solution

7 0

3 years ago