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3 years ago

Find the equation of the linear function represented by the table below in slope-

Mathematics

1 answer:

mixer [17]3 years ago

7 0

Answer:

y= -5x-5

Step-by-step explanation:

because m=y2-y1/x2-x1 and then you plug in one of the point to find the y intercept or b and the answer you should get is -5 for be so it ends up being y= -5x-5

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Can please cross multiply this for me :)

Alchen [17]

64 x 100 = 6400
11 x x = 11x
I’ll solve this out for you
11x = 6400
Divide 11 from both sides
x = 581.8
Hope this helps!

4 0

4 years ago

Read 2 more answers

Is this decreasing or increasing? (1,4) (6,-1)

Nostrana [21]

Answer:

Decreasing

Step-by-step explanation:

It's a linear line going down from left to right

3 0

4 years ago

Increase 400 in the ratio 3:5

lesya [120]

Answer:

250.

Step-by-step explanation:

3 + 5 = 8. => (3/8) * 400 = 3 * 50 = 150. => (5/8) * 400 = 5 * 50 = 250.

4 0

3 years ago

Can someone help me! i struggle with math alot and this one is really hard

fiasKO [112]

Answer:

in this traingle side gh=side hg and angle j is equal to angle k therefore by SAS test the GK siperpendivular to HJ

3 0

3 years ago

Use the tangent identity to

Lunna [17]

Answer:

$tan(x)=\frac{44\sqrt{89}}{89}$

Step-by-step explanation:

tan is defined as: $tan(x)=\frac{opposite}{adjacent}$

sin is defined as: $sin(x)=\frac{opposite}{hypotenuse}$

cos is defined as: $cos(x)=\frac{adjacent}{hypotenuse}$

We can also define tan as: $tan(x)=\frac{sin(x)}{cos(x)}$

because plugging in the definitions of sin and cos in we get: $tan(x)=\frac{(\frac{opposite}{hypotenuse})}{(\frac{adjacent}{hypotenuse})}\\\\tan(x)=\frac{opposite}{hypotenuse}*\frac{hypotenuse}{adjacent}\\\\tan(x)=\frac{opposite}{adjacent}$

which you'll notice is the original definition of tan(x)

So using this definition of tan(x) we can use the givens sin(x) and cos(x) to find tan(x)

$sin(x)=\frac{44}{45}\\\\cos(x)=\frac{\sqrt{89}}{45}\\\\tan(x)=\frac{sin(x)}{cos(x)}$

plugging in sin(x) and cos(x) we get:

$tan(x)=\frac{\frac{44}{45}}{\frac{\sqrt{89}}{45}}\\\\tan(x)=\frac{44}{45}*\frac{45}{\sqrt{89}}\\\\tan(x)=\frac{44}{\sqrt{89}}$

We usually don't like square roots in the denominator, and from here we want to rationalize the denominator which we do by removing the square root from the denominator.

We can do this by multiplying the fraction by: $\frac{\sqrt{89}}{\sqrt{89}}$ which doesn't change the value of the fraction since it simplifies to 1, but it gets rid of the square root in the denominator

$tan(x)=\frac{44}{\sqrt{89}}*\frac{\sqrt{89}}{\sqrt{89}}\\\\tan(x)=\frac{44\sqrt{89}}{89}$

5 0

1 year ago