All categories

3 years ago

Pleasee help me guyss .

Mathematics

2 answers:

Setler79 [48]3 years ago

3 0

Answer:

24%

Step-by-step explanation:

25 - 19 = 6

25 = 100%

2.5 = 10%

1.25 = 5%

0.25 = 1%

5 = 20%

1 = 4%

6 = 24%

Hope this helps :)

natali 33 [55]3 years ago

3 0

Answer:

24% Decrease

Step-by-step explanation:

Start by looking at the formula for percent change.

$\frac{(V^{2} -V^{1})}{V^{1} }$ x 100 = ?

Plug in the numbers, V1 being 25 and V2 being 19.

$\frac{19-25}{25}$ x 100 = ?

19 - 25 = -6

Put -6 as the new numerator.

$\frac{-6}{25}$ x 100 = ?

Divide and get the decimal version of the fraction.

-0.24 x 100 = ?

Multiply and get your final answer of 24% decrease.

You might be interested in

3x+1 <br> can you solve it

Nataly [62]

Answer:

donwload the gautmaht this is very powerful calculator

4 0

2 years ago

Which binomial is a factor of the polynomial? x^2 + x - 6

dlinn [17]

I am guessing the answer is D

4 0

4 years ago

Taylor got 10 out of 12 correct on a proficiency. What percent did she earn?

Lyrx [107]

Answer:

83 percent

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Msr walkers class set up an online fund with a goal to raise 1280

Fofino [41]

Answer: is brainly broken for u too??

Step-by-step explanation:

3 0

3 years ago

Simplify 81p6q2÷3p2q5 HELPPP

goldfiish [28.3K]

Answer:

$\displaystyle \frac{81\, p^{6}\, q^{2}}{3\, p^{2} \, q^{5}} = \frac{27\, p^{4}}{q^{3}}$ .

Step-by-step explanation:

Make use of the fact that for any $x \ne 0$ and integer $n$ :

$\displaystyle \frac{1}{x^{n}} = x^{-n}$ .

For example, in this question:

$\displaystyle \frac{1}{p^{2}} = p^{-2}$ .

$\displaystyle \frac{1}{q^{5}} = q^{-5}$ .

Thus, the original expression would be equivalent to:

$\begin{aligned}& \frac{81\, p^{6}\, q^{2}}{3\, p^{2} \, q^{5}} \\ =\; & \frac{81}{3}\, (p^{6}\, p^{-2})\, (q^{2}\, q^{-5}) \\ =\; & 27\, (p^{6}\, p^{-2})\, (q^{2}\, q^{-5})\end{aligned}$ .

Also make use of the fact that for any $x \ne 0$ , integer $m$ , and integer $n$ :

$x^{m}\, x^{n} = x^{m + n}$ .

Thus:

$\begin{aligned}& 27\, (p^{6}\, p^{-2})\, (q^{2}\, q^{-5}) \\=\; & 27\, p^{6 + (-2)}\, q^{2 + (-5)} \\ =\; & 27\, p^{4}\, q^{-3} \\ =\; & \frac{27\, p^{4}}{q^{3}}\end{aligned}$ .

3 0

2 years ago