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2 years ago

The account charged $35 for the first hour of work and $23 for each hour after that.

Mathematics

1 answer:

defon2 years ago

4 0

Answer:

5 hours

Step-by-step explanation:

127 - 35 = 92

92 / 23 = 4

4 + 1 = 5

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XZ=n, WZ=2x-5, and the perimeter of ΔXWZ is 20. What is the value of n?

Sergio039 [100]

Answer:

Step-by-step explanation:

Since it's an isosceles triangle, WX=WZ= 2n-5

Perimeter = n + (2n-5) + (2n-5)

= 5n - 10

The Perimeter is given to be 20

5n - 10 = 20

5n = 30

n = 6

3 0

2 years ago

Plz help asap with explanation and will get the brilliant

Elina [12.6K]

3x + y = -3
-2x + y = 7
Use elimination method
Subtract both equations
5x = -10, x = -2
Plug in -2 for x (any equation)
3(-2) + y = -3
-6 + y = -3
y = 3

Solution: x = -2, y = 3
Or as a coordinate: (-2, 3)

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2 years ago

Read 2 more answers

I NEED ANSWER ASAP/ WILL GET 20 PTS

ki77a [65]

Answer: A. (-8,-3)

Step-by-step explanation:

Original formula is f(x)=|x-h|+k, the vertex is (h,k), so a positive h in the formula would turn out negative in the vertex.

4 0

3 years ago

Evaluate using <br> Definite integrals

swat32

Since $[0,4]=[0,1]\cup(1,4]$ , we can rewrite the integral as

$\displaystyle \int_0^1f(t)\;dt + \int_1^4 f(t)\; dt$

Now there is no ambiguity about the definition of f(t), because in each integral we are integrating a single part of its piecewise definition:

$\displaystyle \int_0^1f(t)\;dt = \int_0^11-3t^2\;dt,\quad \int_1^4 f(t)\; dt = \int_1^4 2t\; dt$

Both integrals are quite immediate: you only need to use the power rule

$\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}$

to get

$\displaystyle \int_0^11-3t^2\;dt = \left[t-t^3\right]_0^1,\quad \int_1^4 2t\; dt = \left[t^2\right]_1^4$

Now we only need to evaluate the antiderivatives:

$\left[t-t^3\right]_0^1 = 1-1^3=0,\quad \left[t^2\right]_1^4 = 4^2-1^2=15$

So, the final answer is 15.

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3 years ago

Heyyyyy pleasse help me please please and thank you

dimulka [17.4K]

Answer:

x ≈ 33.1°

Step-by-step explanation:

Using the tangent ratio in the right triangle

tan x = $\frac{opposite}{adjacent}$ = $\frac{125}{192}$ , then

x = $tan^{-1}$ ( $\frac{125}{192}$ ) ≈ 33.1° ( to the nearest tenth )

7 0

2 years ago