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3 years ago

100 POINTS!!

Mathematics

2 answers:

jeka57 [31]3 years ago

8 0

Answer:

Answer is B,C, and E

Step-by-step explanation:

I just took the Quiz

Komok [63]3 years ago

3 0

Answer:

its B,C,and E

Step-by-step explanation:

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PLEASE HELP VERY URGENT

worty [1.4K]

Answer:

A. square root of a^2 + b^2 for both answers

Step-by-step explanation:

The first problem, we are given

a^2 + b^2 = c^2

What we do is solve for c.

sqrt(a^2 + b^2) = c

c = sqrt(a^2 + b^2)

For problem 2,

WE can apply the Pythagorean theorem because we have a right triangle.

The equations is

a^2 + b^2 = c^2 like the first problem

Solving gets us

sqrt(a^2 + b^2) = c

c = sqrt(a^2 + b^2)

3 0

1 year ago

Set up an linear equation and solve the following number word problem:

IRINA_888 [86]

Answer:

let the number be y

4(y-7)=48

y-7=48/4

y-7=12

y=12+7=19

I would appreciate if my answer is chosen as a brainliest answer

3 0

2 years ago

Geometric sequences HELP ASAP!

Pani-rosa [81]

Given:

The table for a geometric sequence.

To find:

The formula for the given sequence and the 10th term of the sequence.

Solution:

In the given geometric sequence, the first term is 1120 and the common ratio is:

$r=\dfrac{a_2}{a_1}$

$r=\dfrac{560}{1120}$

$r=0.5$

The nth term of a geometric sequence is:

$a_n=ar^{n-1}$

Where a is the first term and r is the common ratio.

Putting $a=1120, r=0.5$ , we get

$a_n=1120(0.5)^{n-1}$

Therefore, the required formula for the given sequence is $a_n=1120(0.5)^{n-1}$ .

We need to find the 10th term of the given sequence. So, substituting $n=10$ in the above formula.

$a_{10}=1120(0.5)^{10-1}$

$a_{10}=1120(0.5)^{9}$

$a_{10}=1120(0.001953125)$

$a_{10}=2.1875$

Therefore, the 10th term of the given sequence is 2.1875.

6 0

2 years ago

If the two terms of a gemotric sequence are a1=216, and a2=72, which is the third term? a3?

Ierofanga [76]

$GEOMETRIC \: \: PROGRESSIONS \\ \\ \\\\Let \: the \: G.P. \: be \: \: A \: , \: Ar \: , \: A {r}^{2} \: ... \\ \\ Where \:first \: term \: is \: \: A \: \: \\ and \: common \: ratio \: is \: \: R \\ \\ Let \: An \: denotes \: the \: \: nth \: term \: of \: \\ the \: given \: Geometric \: Progression \: \\ \\ It \: is \: given \: - \\ \\ A1 \: = \: 72 \\ \\ A2 \: = \: 216 \\ \\ Common \: ratio \: = \: \frac{A2}{A1} = \frac{216}{72} \\ \\ R = 3 \\ \\ A \: = \: 72 \\ \\ A3 \: = \: A {r}^{2} = 72 \times 3 \times 3 \\ \\ \\ Hence \: , \: A3 \: = \: 648 \: \: \: \: \: \: \: Ans.$