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mrs_skeptik [129]

3 years ago

9

The point slope form of the equation of the line that passes through -5,-1 and 10,-7 is y+7= -2/5(x-10). What is the standard fo

rm of this equation

1 answer:

scoundrel [369]3 years ago

5 0

2x + 5y = -15 is the standard form of equation

<em><u>Solution:</u></em>

Given that,

<em><u>The point slope form of the equation of the line that passes through (-5, -1) and (10, -7) is given by:</u></em>

$y + 7 = \frac{-2}{5}(x-10)$

We have to write the standard form of the equation

The standard form of an equation is Ax + By = C

In this kind of equation, x and y are variables and A, B, and C are integers

<em><u>Let us convert the given equation into standard form</u></em>

$y + 7 = \frac{-2}{5}(x-10)\\\\\text{Simplify the right hand side of equation }\\\\y + 7 = \frac{-2x}{5}+\frac{20}{5}\\\\y + 7 = \frac{-2x +20}{5}\\\\5(y+7) = -2x + 20\\\\\text{Simplify the left hand side of equation }\\\\5y + 35 = -2x + 20\\\\\text{Move the variables to left side of equation and constants to right side }\\\\2x + 5y = 20 - 35\\\\2x + 5y = -15$

Thus the standard form of equation is found

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Ten workers are hired to harvest potatoes from a field. Each is givin a plot which is 6x11 feet size. What is the total area of

Monica [59]

Answer:

66

Step-by-step explanation:

The answer is 66 because 6 x 11 equals 66!!!!

Do Mental math if not work on it on a piece of paper.

6 0

3 years ago

Find a linear approximation of the function f(x)=\sqrt[3]{1+x} at a=0, and use it to approximate the numbers \sqrt[3]{.96} and \

34kurt

$f(x)=\sqrt[3]{1+x}=(1+x)^{1/3}\implies f'(x)=\dfrac1{3(1+x)^{2/3}}$

The linear approximation to $f(x)$ around $x=a$ is then

$L(x)=f(a)+f'(a)(x-a)\approx f(x)$

So the approximation centered at $a=0$ will be

$L(x)=f(0)+f'(0)x=(1+0)^{1/3}+\dfrac x{3(1+0)^{2/3}}$
$L(x)=1+\dfrac x3$

which means we have

$\sqrt[3]{0.96}\approx L(0.96)=1+\dfrac{0.96}3=1.32$

$\sqrt[3]{1.02}\approx L(1.02)=1+\dfrac{1.02}3=1.34$

Compare to the actual values of

$\sqrt[3]{0.96}\approx0.9864$

$\sqrt[3]{1.02}\approx1.0066$

4 0

3 years ago

1-cos(6x)=___?<br> A. 3sin(2x)<br> B. 2sin^2(3x)<br> C. 3cos(2x)<br> D. 2cos^2(3x)

zhenek [66]

The solution to 1 - cos(6x) is 2sin²(x).

Hence, option B) 2sin²(x) is the correct answer.

<h3>What is solution to 1 - cos(6x)?</h3>

Given that; 1 - cos(6x) = ?

First, we rewrite using trig identity

1 - cos(2 × 3x)

Using the double angle identity, { <em>cos2(x) = 1 - 2sin²(x)</em> }

1 - ( 1 - 2sin²(x) )

Eliminate the parentheses

1 - 1 + 2sin²(x)

2sin²(x)

The solution to 1 - cos(6x) is 2sin²(x).

Hence, option B) 2sin²(x) is the correct answer.

Learn more about trig identity here: brainly.com/question/10376944

#SPJ1

4 0

2 years ago

Why are there two solutions for the equation |6 + y| = 2? Explain.

GaryK [48]

$Solution, \left|6+y\right|=2\quad :\quad y=-8\quad \mathrm{or}\quad \:y=-4$

$Steps:$

$|f\left(y\right)|=a\quad \Rightarrow \:f\left(y\right)=-a\quad \mathrm{or}\quad \:f\left(y\right)=a, 6+y=-2\quad \quad \mathrm{or}\quad \:\quad \:6+y=2$

$6+y=-2\quad :\quad y=-8,\\6+y=-2,\\\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides}, 6+y-6=-2-6,\\\mathrm{Simplify}, y=-8$

$6+y=2\quad :\quad y=-4,\\6+y=2,\\\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides},6+y-6=2-6\\\mathrm{Simplify},y=-4$

$\mathrm{Combine\:the\:ranges}, y=-8\quad \mathrm{or}\quad \:y=-4$

$\mathrm{The\:Correct\:Answer\:is\:Because\:of\:the\:absolute\:value,\:It\:could\:be\:Positive\:or\:negative.}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$

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4 years ago

Use the diagram below to find the solution to -3 (3)/(2)+2

lisabon 2012 [21]

Answer:

wheres the diagram? wondering if it just didnt copy.

Step-by-step explanation:

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4 years ago