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Alex Ar [27]
3 years ago
7

What’s x Pythagorean triples

Mathematics
1 answer:
nadezda [96]3 years ago
6 0
We have to find x and x is the hypotenuse. We use this formula:

hyp²=side²+side²

So:

hyp²=side²+side²

hyp²= 12²+9²

hyp²= 225

hyp²= √225

hyp= 15

Therefore, the hypotenuse is 15 and x is 15.
HOPE THIS HELPS!
HAVE A GR8 EVENING ;-)

You might be interested in
Work out the percentage change to 2 decimal places when a price of £57.99 is decreased to £49.99.
IRISSAK [1]

Answer:

Percentage change in price = 13.80%

Step-by-step explanation:

Given that:

Original price = £57.99

Decreased price = £49.99

Difference = Original price - Decreased price

Difference = 57.99 - 49.99

Difference = £8.00

Percent change = \frac{Difference}{Original\ price}100

Percent change = \frac{8.00}{57.99}*100

Percent change in price = 13.80%

Hence,

Percentage change = 13.80%

5 0
3 years ago
Solve for x.<br><br> x2−10=−5<br><br><br><br> x=±5√<br><br> x=±15−−√<br><br> x=±35√<br><br> x=±50−−√
aksik [14]

Answer:

x=±5√

hope this helped

Brainliest?

5 0
2 years ago
Read 2 more answers
Someone please HELP!?
jeka94
 A negative exponent just means that the base is on the wrong side of the fraction line, so you need to flip the base to the other side.

For example

x^{-2}=\frac{1}{x^2}

or

( \frac{2}{5} )^{-4}=( \frac{5}{2} )^4

***************************************
(8r^{-5})^{-3}

(8* \frac{1}{r^5} )^{-3} \\\\( \frac{8}{r^5} )^{-3}&#10; \\  \\ ( \frac{r^5}{8} )^3&#10;\\\\ \mathrm{Apply\:exponent\:rule}:\quad *\left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} \ \ \ \ \ \ \ \ \ \ \ \   *(a^b)^c=a^{bc}&#10; \\\\( \frac{r^{15}}{8^3} )&#10;\\\\( \frac{r^{15}}{512} )

The answer is "D"


7 0
2 years ago
A number increased by 5 is two times the number
marin [14]
Let x be the number. x + 5 = 2x. 5 = x because you isolate for x by moving it to the right side, meaning you’d have to subtracts 1 x from 2x
8 0
3 years ago
Suppose that the functions r and a are defined for all real numbers x as follows. r(x)=2x-1 S(x)=5x write the expressions for (r
NeTakaya

\boxed{(r-s)(x)=-3x-1} \\ \\ \boxed{(r\cdot s)(x)=10x^2-5x} \\ \\ \boxed{(r+s)(-2)=-15}

<h2>Explanation:</h2>

In this exercise, we have the following functions:

r(x)=2x-1 \\ \\ s(x)=5x

And they are defined for all real numbers x. So we have to write the following expressions:

First expression:

(r-s)(x)

That is, we subtract s(x) from r(x):

(r-s)(x)=2x-1-5x \\ \\ Combine \ like \ terms: \\ \\ (r-s)(x)=(2x-5x)-1 \\ \\ \boxed{(r-s)(x)=-3x-1}

Second expression:

(r\cdot s)(x)

That is, we get the product of s(x) and r(x):

(r\cdot s)(x)=(2x-1)(5x) \\ \\ By \ distributive \ property: \\ \\ (r\cdot s)(x)=(2x)(5x)-(1)(5x) \\ \\ \boxed{(r\cdot s)(x)=10x^2-5x}

Third expression:

Here we need to evaluate:

(r+s)(-2)

First of all, we find the sum of functions r(x) and s(x):

(r+s)(x)=2x-1+5x \\ \\ Combine \ like \ terms: \\ \\ (r+s)(x)=(2x+5x)-1 \\ \\ (r+s)(x)=7x-1

Finally, substituting x = -2:

(r+s)(-2)=7(-2)-1 \\ \\ (r+s)(-2)=-14-1 \\ \\ \boxed{(r+s)(-2)=-15}

<h2>Learn more: </h2>

Parabola: brainly.com/question/12178203

#LearnWithBrainly

5 0
3 years ago
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