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2 years ago

Writing equations in standard form:

1 answer:

3 0

You have already done half of it. Now substitute for x1 = 4 and y = 5
y - 5 = 6/7(x - 4)
Multiply everything by 7 to remove fractions
7y - 35 = 6(x - 4)
7y - 35 = 6x - 24
-6x + 7y = -24 + 35
-6x + 7y = 11

Some people prefer a positive number in front of x so multiply by -1
6x - 7y = -11

You can check the answer by testing it works for both of the original points
6x4 - 7x5 = 24 - 35 = -11 true
6x-3 - 7x-1 = -18 - -7 = -11 true

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ANSWER PLEASE

SSSSS [86.1K]

Answer:

5/8

Step-by-step explanation:

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3 years ago

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Which of the following inequalities matches the graph?

S_A_V [24]

X<-3 would be the answer

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2 years ago

QUIZ

yanalaym [24]

Y = x + 2

It starts out at (0,2) which means it needs to be up 2 from the parent graph (y = x). Then using the slope equation, the slope is 1

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3 years ago

Try to sketch by hand the curve of intersection of the parabolic cylinder y = x2 and the top half of the ellipsoid x2 + 7y2 + 7z

vovikov84 [41]

Plug $y=x^2$ into the equation of the ellipsoid:

$x^2+7(x^2)^2+7z^2=49$

Complete the square:

$7x^4+x^2=7\left(x^4+\dfrac{x^2}7+\dfrac1{14^2}-\dfrac1{14^2}\right)=7\left(x^2+\dfrac1{14}\right)^2+\dfrac1{28}$

Then the intersection is such that

$7\left(x^2+\dfrac1{14}\right)^2+7z^2=\dfrac{1371}{28}$

$\left(x^2+\dfrac1{14}\right)^2+z^2=\dfrac{1371}{196}$

which resembles the equation of a circle, and suggests a parameterization is polar-like coordinates. Let

$x(t)^2+\dfrac1{14}=\sqrt{\dfrac{1371}{196}}\cos t\implies x(t)=\pm\sqrt{\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}}$

$y(t)=x(t)^2=\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}$

$z=\sqrt{\dfrac{1371}{196}}\sin t$

(Attached is a plot of the two surfaces and the intersection; red for the positive root $x(t)$ , blue for the negative)

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3 years ago

Evaluate: 6 + [4(5 + 9)] step by step

jek_recluse [69]

The answer is 62
First you would have to add the ones in the parentheses 5+9=14 then you would have to multiply it by the 4 [4 14]giving you 56. Then 56+6= 62

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3 years ago