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3 years ago

What is the system of equations y=-3x+7 y=2x-8

Mathematics

2 answers:

frez [133]3 years ago

6 0

Answer:(6,4)

Step-by-step explanation: y=4

y=2x-8

4=2x-8

X=6

Y=4

sveta [45]3 years ago

6 0

(6,4) I think I’m sorry if this not correct

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Which graph represents the linear function y = x 3/4– 4?

ahrayia [7]

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

$y=\frac{3}{4}x-4$

This is a linear equation (the graph is a line)

To identify the graph find out the intercepts

Find out the y-intercept

The y-intercept is the value of y when the value of x is equal to zero

For x=0

$y=\frac{3}{4}(0)-4=-4$

The y-intercept is the point (0,-4)

Find out the x-intercept

The x-intercept is the value of x when the value of y is equal to zero

For y=0

$0=\frac{3}{4}x-4$

$\frac{3}{4}x=4$

$x=16/3=5.33$

The x-intercept is the point (5.33,0)

therefore

The graph in the attached figure

3 0

3 years ago

4 + 8x = -12 what is X?

Elena L [17]

Answer:

x=-2

Step-by-step explanation:

4+8x=-12

-4 -4

8x = -16

/8 /8

x = -2

5 0

3 years ago

Read 2 more answers

Bianca wrote the steps for her solution to the equation 2.3 + 8(1.3x – 4.75) = 629.9. She left out the description for the last

Virty [35]

I believe dividing is always the last step.

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3 years ago

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Find an equation of the line that has the given slope and contains the given point. m=7/3 (1,6)

iris [78.8K]

The point-slope form:

$y-y_1=m(x-x_1)$

We have the point (1, 6) and the slope m = 7/3. Substitute:

$y-6=\dfrac{7}{3}(x-1)$ use distributive property

$y-6=\dfrac{7}{3}x-\dfrac{7}{3}$ add 6 to both sides

$y=\dfrac{7}{3}x+\dfrac{11}{3}$ multiply both sides by 3

$3y=7x+11$ subtract 7x from both sides

$-7x+3y=11$ change the signs

$7x-3y=-11$

Answer:

point-slope form: $y-6=\dfrac{7}{3}(x-1)$

slope-intercept form: $y=\dfrac{7}{3}x+\dfrac{11}{3}$

standard form: $7x-3y=-11$

5 0

3 years ago

Prove that 1+cosA/sinA + sinA/1+cosA=2cosecA

Alla [95]

$\frac{1+cos\alpha}{sin\alpha}+\frac{sin\alpha}{1+cos\alpha}=2cosec\alpha\\\\L=\frac{(1+cos\alpha)(1+cos\alpha)+sin\alpha\cdot sin\alpha}{sin\alpha(1+cos\alpha)}=\frac{1+2cos\alpha+cos^2\alpha+sin^2\alpha}{sin\alpha(1+cos\alpha)}\\\\=\frac{1+2cos\alpha+1}{sin\alpha(1+cos\alpha)}=\frac{2+2cos\alpha}{sin\alpha(1+cos\alpha)}=\frac{2(1+cos\alpha)}{sin\alpha(1+cos\alpha)}\\\\=\frac{2}{sin\alpha}=2\cdot\frac{1}{sin\alpha}=2cosec\alpha=R$

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3 years ago