All categories

3 years ago

Please help me guys :(((((((

Mathematics

1 answer:

zysi [14]3 years ago

3 0

Your answer is A its the only one where it subtracts the original price with the discount

You might be interested in

If the sum of 1 and the product<br> of p and 8 is equal to 62 less than p

Ierofanga [76]

Uhh I think isa carrot

5 0

3 years ago

Geometry work pls help ASAP

ki77a [65]

Answer:

3321

Step-by-step explanation:

6 0

2 years ago

SOMEONE PLEASE HELP ME!!!!! I NEED THIS!

snow_tiger [21]

Answer:

QR ≈ 16.3

Step-by-step explanation:

Since the figures are similar then the ratios of corresponding sides are equal, that is

$\frac{MN}{QR}$ = $\frac{MP}{QT}$ , substitute values

$\frac{40.3}{QR}$ = $\frac{57}{23}$ ( cross- multiply )

57 QR = 926.9 ( divide both sides by 57 )

QR ≈ 16.3 ( to the nearest tenth )

7 0

3 years ago

The sum of 2 times a number and 10 is the same as 6 times the difference of the number and 1. What is the number?

likoan [24]

Answer:

Step-by-step explanation:

if the required number is 'x', then the sum of 2 times the number and 10 is 2x+10; the difference of the number and 1 is x-1. Using these two equation it is possible to make up the common one:

2x+10=6(x-1);

4x=16;

x=4.

7 0

3 years ago

Simply (m^3np^-3)^3(mn^4p^2)

Tamiku [17]

Answer:

$=\frac{m^{10}n^7}{p^7}$

Step-by-step explanation:

$(m^3np^-^3)^3(mn^4p^2)\\\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=\left(m^3np^{-3}\right)^3mn^4p^2\\\\\left(m^3np^{-3}\right)^3\quad :\quad m^9n^3p^{-9}\\\\=m^9n^3p^{-9}mn^4p^2\\\\\mathrm{Apply\:exponent\:rule}:\quad \:a^b\times\:a^c=a^{b+c}\\\\m^9m=\:m^{9+1}\\\\=n^3p^{-9}m^{9+1}n^4p^2\\n^3n^4=\:n^{3+4}\\\\=n^3p^{-9}m^{10}n^4p^2\\\\\mathrm{Apply\:exponent\:rule}:\quad \:a^b\times\:a^c=a^{b+c}\\=p^{-9}m^{10}n^{3+4}p^2\\\\=p^{-9}m^{10}n^7p^2\\\\=m^{10}n^7p^{-9+2}$

$=m^{10}n^7p^{-7}\\\\=m^{10}n^7\frac{1}{p^7}\\\\\frac{1\times\:m^{10}n^7}{p^7}\\\\=\frac{m^{10}n^7}{p^7}$

4 0

3 years ago