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3 years ago

I need help with these questions

Mathematics

1 answer:

otez555 [7]3 years ago

7 0

1) First you would multiply 3 × -4 which gives you -12

You add exponents when multiplying.

So, x^2 +x^3 = x^5 and y^3 +y^4 = y^7

Final Answer: -12x^5y^7

2) 7 times 4 is 28 and y^3 +y^5 = y^8

Final Answer: 28y^8

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Given f ( x ) = 1 x + 10 , find the average rate of change of f ( x ) on the interval [ 8 , 8 + h ] . Your answer will be an exp

son4ous [18]

Answer: 1

Step-by-step explanation:

$f(8+h)=8+h+10=18+h\\\\f(8)=8+10=18$

So, the average rate of change is:

$\frac{(18+h)-18}{(8+h)-8}=\frac{h}{h}=\boxed{1}$

The question is wrong, the expression will not be in terms of h since it is a linear function.

6 0

2 years ago

I need help with this

maksim [4K]

Answer:

The answer is B

Step-by-step explanation:

6 0

3 years ago

In ΔWXY, the measure of ∠Y=90°, XW = 53, YX = 28, and WY = 45. What is the value of the cosine of ∠X to the nearest hundredth?

kotegsom [21]

The value of ∠X = 58.11°, If ΔWXY, the measure of ∠Y=90°, XW = 53, YX = 28, and WY = 45.

Step-by-step explanation:

The given is,

In ΔWXY, ∠Y=90°

XW = 53

YX = 28

WY = 45

Step:1

Ref the attachment,

Given triangle XWY is right angled triangle.

Trigonometric ratio's,

$Cos$ ∅ $= \frac{Adj}{Hyp}$

For the given attachment, the trigonometric ratio becomes,

$Cos$ ∅ $= \frac{XY}{XW}$ .....................................(1)

Let, ∠X = ∅

Where, XY = 28

XW = 53

Equation (1) becomes,

$Cos$ ∅ $= \frac{28}{53}$

$Cos$ ∅ = 0.5283

∅ = $cos^{-1}$ (0.5283)

∅ = 58.109°

Result:

The value of ∠X = 58.11°, If ΔWXY, the measure of ∠Y=90°, XW = 53, YX = 28, and WY = 45.

4 0

3 years ago

The center of an ice rink is located at (0, 0) on a coordinate system measured in feet. sandi skates along a path that can be mo

WARRIOR [948]

The answers to the question are are the equations

y = 0.08x²-1.6x+13

y = 0.08(x-10)²+5

<h3>How to solve for the system of equations</h3>

David starts at (20, 13) and skates along a path that can be modeled by a quadratic function with a vertex at (10, 5).

Vertex form of quadratic function is

y = $a(x-h)^2+k$

13 = a(x-10)²+5

let x = 20

a(20-10)²+5

13 = 100a + 5

take like terms

13 - 5 = 100a

8 = 100a

divide through by 100

a = 0.08

hence the equation would be

y = 0.08(x-10)²+5