Help me plsx xndnjdsjjs

Mathematics

1 answer:

Kipish [7]3 years ago

6 0

There is the answer hope it helpd if you can mark me brainliest

You might be interested in

What are three different ways to write 5 to the 11 power

Olin [163]

(5)(5)(5)(5)(5)(5)(5)(5)(5)(5)(5)
5^11
That is two of the three but I'm not quite sure if there is a third.

3 0

3 years ago

Read 2 more answers

HOTELS At the Eastward Inn hotel, 47% of the rooms face the pool. The hotel has 92 rooms. Estimate the number of rooms that face

denis-greek [22]

About 43 roomsAbout 43 roomsvAbout 43 roomsAbout 43 roomsAbout 43 roomsAbout 43 roomsAbout 43 roomsAbout 43 roomsAbout 43 roomsAbout 43 roomsAbout 43 roomsAbout 43 rooms

4 0

3 years ago

This graph represents a transformation of the parent cube root function. Replace the values of h and k to create the equation of

Zigmanuir [339]

Answer: H is 5 and K is 2

Step-by-step explanation:

8 0

3 years ago

Which expression is equivalent to (r Superscript negative 7 Baseline) Superscript 6? r Superscript 42 StartFraction 1 Over r Sup

vodomira [7]

Power rule of exponents says that the power of the power of a exponent is equal to the multiplying both the powers. The expression which is equal to the given expression is,

$\dfrac{1}{(r)^{42}}$

The option B is the correct option.

Given information-

The given expression in the problem is,

$[(r)^{-7}]^5$

<h3>Power rule of exponents</h3>

Power rule of exponents says that the power of the power of a exponent is equal to the multiplying both the powers.

Suppose<em> </em><em>x</em> is a number with power <em>a. </em>The power is power of the power of number <em>x. </em>Then by the power rule of the exponents,

$(x^a)^b=x^{ab}$

Using the power rule of the exponents given expression can be written as,

$[(r)^{-7}]^6=(r)^{-7\times 6} 
$

$[(r)^{-7}]^6=(r)^{-42}$

<h3>Negative exponents rule</h3>

Negative exponents rule says that when a power of a number is negative, then write the number in the denominator with the same power with positive sign.Thus,

$[(r)^{-7}]^6=\dfrac{1}{(r)^{42}}$