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3 years ago

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 ;-)

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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

3 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

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

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

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

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

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