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Morgarella [4.7K]
3 years ago
7

What is the solution to the inequality -2(m+3)<4​

Mathematics
1 answer:
diamong [38]3 years ago
6 0

Answer:

m>−5

Step-by-step explanation:

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What is the standard form equation of the line shown below?
Pavlova-9 [17]

Answer:

C.

Step-by-step explanation:

14/3 = 4.7 which seems about right for the Y-intercept & the gradient is - 1/3.

Gradient = rise /run

= 1/3

Hope this helps!

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2 years ago
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Find x A. 212√2 B. 212–√ C. 213√2 D. 7
victus00 [196]

The value of x in the triangle is (d) 7

<h3>How to solve for x?</h3>

The complete question is in the attached image.

From the attached image of the triangle, we can see that the triangle is a right triangle, and x can be solved using the following sine function

\sin(45) = \frac{x}{7\sqrt 2}

Evaluate sin(45)

\frac 1{\sqrt 2} = \frac{x}{7\sqrt 2}

Solve for x

x = \frac{7\sqrt 2}{\sqrt 2}

Divide

x = 7

Hence, the value of x in the triangle is (d) 7

Read more about special triangles at:

brainly.com/question/654982

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7 0
2 years ago
A school typically sells 500 yearbooks in a year for $50 dollars each.
Evgen [1.6K]
You divide 500 by $50 to get 10; the school sold 10 yearbooks that year if you’re looking for that as an answer.
8 0
3 years ago
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The surface area of a cone is found using the formula SA = Pir2 + Pirl. Describe what each part of the formula represents and ho
Elza [17]

Answer:

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3 years ago
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Find the area bounded by the graphs of the indicated equations over the given interval.
NeX [460]

Answer:

48

Step-by-step explanation:

y = 0

is basically the horizontal axis.

First, find the integral of x^2-25.

Remember that

integral of a constant is that constant times x.

Also that

to take the integral of a power function, add 1 to the degree and divide by that same degree.

y =  {x}^{2}  - 25

We then get

\frac{ {x}^{3} }{3}  - 25x

Evaluate at -3

\frac{ - 3 {}^{3} }{3}  - 25( - 3)

- 27 + 75 = 48

Then we evaluate at 0

\frac{0 {}^{3} }{3}  - 25(0) = 0

Next, we subtract the the answer then we get

48 - 0 = 48

8 0
2 years ago
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