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GrogVix [38]
3 years ago
7

Hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

Mathematics
2 answers:
Tom [10]3 years ago
8 0

Answer:

same

Step-by-step explanation:

35 h's you typed

navik [9.2K]3 years ago
6 0

Answer:

I see, you have to hhh.

Step-by-step explanation:

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A wildlife reserve had 8 elephant calves born during the summer and now has 31 total elephants. How many elephants were reserved
julia-pushkina [17]
The correct answer is 23

5 0
2 years ago
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Express 96 as the product of prime factors<br>need it asap<br>plzz write the method too​
Murrr4er [49]

Given:

The given number is:

96

To find:

The product of prime factors of given number.

Solution:

We have,

96

The prime factors of given number are:

<u>2 | 96</u>

<u>2 | 48</u>

<u>2 | 24</u>

<u>2 | 12</u>

<u>2 | 6</u>

<u>3 | 3</u>

  | 1    

So, 96 can be written as:

96=2\times 2\times 2\times 2\times 2\times 3

Therefore, the 96 as the product of prime factors is 96=2\times 2\times 2\times 2\times 2\times 3.

3 0
2 years ago
What are the points for f(x)=3^(x-1)-2
Savatey [412]

Answer:  (0, -1 \frac{2}{3}), (1, -1), (2, 1)

<u>Step-by-step explanation:</u>

Set the exponent equal to zero - <em>that is your anchor point</em>.

Then choose an x-value less than and greater than the anchor point.

\left\begin{array}{c|c|l}&\underline{\quad x\quad }&\underline{\qquad \qquad f(x)\qquad \qquad \qquad \quad }\\\text{less than anchor}&0&3^{0-1}-2=3^{-1}-2=\frac{1}{3}-2 = -1\frac{2}{3}}\\\\anchor\ point&1&3^{1-1}-2=3^{0}-2=1-2 = -1}\\\\\text{greater than anchor}&2&3^{2-1}-2=3^{1}-2=3-2 = 1}\end{array}\right

4 0
3 years ago
I dont know how to do these questions I would be pleased if someone helped thankyou.
sleet_krkn [62]

Area = area of the square in the center + 2 * area of the 2 quarter circles.

= 5^2 + 2 * 1/4 *3.14 *  5^2

= 25 + 39.25

= 64.3 in^2    answer

5 0
3 years ago
Which of the following can be a Pythagorean triplets?:
PolarNik [594]

<h2>Answer :-</h2>

As we know that,

Pythagoras triplet

1) a² + b² = c²

Let

\sf \: {2}^{2} + {2}^{2} = {4}^{2}

\sf \: 4 + 4 = 16

  • 8 ≠ 16
<h3>Hence, A can't be Pythagoras triplet</h3>

2) a² + b² = c²

\sf \:5 {}^{2} + {12}^{2} = {13}^{2}

\sf \: 25 + 144 = 169

  • 169 = 169
<h3>Therefore, B can be Pythagoras triplet</h3>

3)a² + b² = c²

\sf {3}^{2} + {5}^{2} = {6}^{2}

\sf 9 + 25 = 36

  • 34 ≠ 36
<h3>Hence, C can't be Pythagoras triplet</h3>

4) a² + b² = c²

\sf {5}^{2} + {4}^{2} = {7}^{2}

\sf \: 25 + 16 = 49

  • 41 ≠ 49
<h3>Hence, D can't be Pythagoras triplet</h3>

<h2 /><h2>Therefore :-</h2>

Only B can be Pythagoras triplet.

3 0
3 years ago
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