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Sedbober [7]
3 years ago
9

How can I find c in a=3 b=4

Mathematics
1 answer:
Dahasolnce [82]3 years ago
8 0
A^2+b^2= c^2
3^2+4^2=c^2
9+16=c^2
25=c^2
5=c
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In ΔEFG, e = 8.9 cm, f = 7.7 cm and g=2.2 cm. Find the measure of ∠F to the nearest degree.
Setler79 [48]

Answer:

51 degrees

Step-by-step explanation:

gave up on delta math

7 0
3 years ago
If the sum of $1! + 2! + 3! + \cdots + 49! + 50!$ is divided by $15$, what is the remainder?
kompoz [17]

Answer:

  3

Step-by-step explanation:

All factorials 5 and above are evenly divisible by 15, so have no remainder. Thus, you are interested in ...

  mod(1! +2! +3! +4!, 15) = mod(1 +2 +6 +24, 15)

  = mod(33, 15) = 3

The remainder is 3.

3 0
3 years ago
Idk if this is correct or not !!!!!!!! WILL MARK BRIANLIEST !!!!!!!! PLEASE HELP !!!!!!!!!!!!!!!
andrew-mc [135]

Answer:

yes i took this test thats correct ! brainliest please :)

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
John is 14 years older than Sam and the product of their ages is 240. How old is Sam?
Shtirlitz [24]

Answer:

A.) 10 years old

Step-by-step explanation:

Sam: John - 14

S = J - 14

S x J = 240

10 = J - 14

0 = J - 14 - 10

-J = -24

J = 24

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3 0
3 years ago
Given the following equation of a circle in general form, find the equation in standard form by completing the square.
Artist 52 [7]

The standard form of the circle equation 4x² + 8x + 4y² + 32y +52 = 0 is (x + 1)² + (y + 4)² = 2²

<h3>What is a circle?</h3>

It is described as a set of points, where each point is at the same distance from a fixed point (called the center of a circle)

We have an equation that represents the circle:

4x² + 8x + 4y² + 32y +52 = 0

Divide by 4 on both the sides:

x² + 2x + y² + 8y + 13 = 0

x² + 2x + 1 - 1 + y² + 8y + 4² - 4² + 13 = 0

x² + 2x + 1 + y² + 8y + 4² - 1 - 16 + 13

(x + 1)² + (y + 4)² = 4

(x + 1)² + (y + 4)² = 2²

Thus, the standard form of the circle equation 4x² + 8x + 4y² + 32y +52 = 0 is (x + 1)² + (y + 4)² = 2²

Learn more about circle here:

brainly.com/question/11833983

#SPJ1

3 0
2 years ago
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