Find the cube of five

I need help I dont understand

svp [43]

Answer:

X=5

mde=e

mef=5

mdfg=77

Step-by-step explanation:

6 0

3 years ago

Can someon help me plis?

PtichkaEL [24]

this one lol

Step-by-step explanation:

because this shape has 12 sides when every other shape has 13

4 0

3 years ago

Read 2 more answers

It is estimated that there are 10^20 grains of sand on a beach in Florida in April. The beach is heavily eroded by tourists over

lyudmila [28]

$\bf \textit{how much is }\frac{1}{100}\quad of \quad 10^{20}?\implies \stackrel{\textit{one hundredth}}{\cfrac{1}{100}}\cdot 10^{20}\implies \cfrac{1}{10^2}\cdot 10^{20}
\\\\\\
10^{-2}\cdot 10^{20}\implies 10^{-2+20}\implies 10^{18}
\\\\\\
1,000,000,000,000,000,000~grains$

5 0

3 years ago

The coordinates of point A are (p, q) and coordinates of point B are (p+2q, q+2p). Provide your complete solutions and proofs in

Dafna1 [17]

Answer:

The mid-point (p + q , q + p) of AB is the same distance from the x-axis and the y-axis

Step-by-step explanation:

* Lets explain how to solve the problem

- Any point will be equidistant from the x-axis and the y-axis must have

equal coordinates

- Ex: point (4 , 4) is the same distance from the x-axis and the y-axis

because the distance from the x-axis to the point is 4 (y-coordinate)

and the distance from the y-axis and the point is 4 (x-coordinate)

- If (x , y) is the mid-point of a segment its endpoints are

$(x_{1},y_{1})$ and $(x_{2},y_{2})$ , then

$x=\frac{x_{1}+x_{2}}{2}$ and $y=\frac{y_{1}+y_{2}}{2}$

* Lets solve the problem

∵ Point A has coordinates (p , q)

∵ Point B has coordinates (p + 2q , q + 2p)

- The mid-point of AB is (x , y)

∵ $x=\frac{p+p+2q}{2}$

∴ $x=\frac{2p+2q}{2}$

- Take 2 as a common factor from the terms of the numerator

∴ $x=\frac{2(p+q)}{2}$

- Divide up and down by 2

∴ x = p + q

∵ $y=\frac{q+q+2p}{2}$

∴ $y=\frac{2q+2p}{2}$

- Take 2 as a common factor from the terms of the numerator

∴ $y=\frac{2(q+p)}{2}$

- Divide up and down by 2

∴ y = q + p

∴ The mid point of AB is (p + q , q + p)

- p + q is the same with q + p

∵ The x-coordinate of the mid point of AB is p + q

∵ The y-coordinate of the mid point of AB is q + p

∵ p + q = q + p

∴ The coordinates of the mid-point of AB are equal

- According the explanation above

∴ The mid-point (p + q , q + p) of AB is the same distance from the

x-axis and the y-axis

8 0

3 years ago

Simplify (12a5−6a−10a3)−(10a−2a5−14a4) . Write the answer in standard form

almond37 [142]

Answer:

$=14a^5+14a^4-10a^3-16a$

Step-by-step explanation:

$\left(12a^5-6a-10a^3\right)-\left(10a-2a^5-14a^4\right)\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=12a^5-6a-10a^3-\left(10a-2a^5-14a^4\right)\\-\left(10a-2a^5-14a^4\right):\quad -10a+2a^5+14a^4\\-\left(10a-2a^5-14a^4\right)\\\mathrm{Distribute\:parentheses}\\=-\left(10a\right)-\left(-2a^5\right)-\left(-14a^4\right)\\Apply\:minus-plus\:rules\\-\left(-a\right)=a,\:\:\:-\left(a\right)=-a\\=-10a+2a^5+14a^4\\=12a^5-6a-10a^3-10a+2a^5+14a^4$

$\mathrm{Simplify}\:12a^5-6a-10a^3-10a+2a^5+14a^4:\quad 14a^5+14a^4-10a^3-16a12a^5-6a-10a^3-10a+2a^5+14a^4\\Group\:like\:terms\\=12a^5+2a^5+14a^4-10a^3-6a-10a\\\mathrm{Add\:similar\:elements:}\:12a^5+2a^5=14a^5\\=14a^5+14a^4-10a^3-6a-10a\\\mathrm{Add\:similar\:elements:}\:-6a-10a=-16a\\=14a^5+14a^4-10a^3-16a$

4 0

3 years ago

Find the cube of five​

Find the cube of five