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Delvig [45]
3 years ago
13

PLEASE HELP ME AND DONT JOKE AROUND

Mathematics
2 answers:
Nuetrik [128]3 years ago
6 0
The answer is 12 hope this helped
VladimirAG [237]3 years ago
5 0

Answer:

12

Step-by-step explanation:

Simplify  2+3 to 5.

7+3×  3 /5

Cancel 3

7+5

Simplify.

12

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Help me pleaseeeeeeeee
kirill [66]

Answer:

B. 3.45

Step-by-step explanation:

The radius is half the diameter

8 0
3 years ago
Read 2 more answers
Lyle uses a streaming service to watch movies. Last month he paid $26.91 and watched 9 movies. If he watches 11 movies this mont
Anastasy [175]

Answer:

A. $32.89

Step-by-step explanation:

Take the amount and divide by the number of movies watched

26.91 / 9 =2.99

Now multiply by the new number of movies he wants to watch, 11

2,99 *11

32.89

5 0
11 months ago
Read 2 more answers
What would my GPA be? Here are my grades.
marta [7]
That is approximately, in order, a C, a C, a B, a D, an A, a B, and an A.

In grade points that is a 2, a 2, a 3, a 1, a 4, a 3, and a 4.

The average of those numbers is about 2.7, so you have a 2.7.

You can raise that by bringing up the D as it is an outlier here.
4 0
3 years ago
Line m is parallel to line p. m ∠ HEF = 39º and m ∠ IGF = 13º. Find the m ∠ EFG. Explain in detail how you know you are correct.
svetlana [45]

Answer:

The measure of ∠EFG is 52°

Step-by-step explanation:

Given line m is parallel to line p. m∠HEF = 39º and m∠IGF = 13º.we have to find m∠EFG.

In ΔJFG,

By angle sum property of triangle, which states that sum of all angles of triangle is 180°

m∠FJG+m∠JGF+m∠JFG=180°

⇒ 39°+13°+m∠JFG=180°

⇒ m∠JFG=180°-39°-13°=128°

As JFE is a straight line ∴ ∠JFG and ∠EFG forms linear pair

⇒ m∠JFG+m∠EFG=180°

⇒ 128°+m∠EFG=180°

⇒ m∠EFG=52°

The measure of ∠EFG is 52°

6 0
3 years ago
Please help! I attached the question below.
kompoz [17]

Answer:

\frac{2(c+2)}{c(c-2)}

Step-by-step explanation:

\frac{c^{2}-4 }{6c^{4}+15c^{3}}=\frac{(c-2)(c+2)}{c(6c^{3}+15c^{2}) }

Identity used:

a^{2}-b^{2}=(a-b)(a+b)

\frac{c^{2}-4c+4}{12c^{3}+30c^{2}}=\frac{(c-2)^{2}}{2(6c^{3}+15c^{2}) }

Now let us divide the modified expressions:

\frac{(c-2)(c+2)}{c(6c^{3}+15c^{2})} ÷ \frac{(c-2)^2}{2(6c^{3}+15c^{2}) }

we get:

\frac{2(c+2)}{c(c-2)}

5 0
3 years ago
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