Simplify. (4^3)5 a. 4^-2 b. 4^3 c. 4^8 d. 4^15

What number do 7 6 3 and 5 go into(lcm)

laila [671]

Answer:

210

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Sally makes $9.15 per hour. If she works 3.5 hours, how much money will she make?

liraira [26]

Answer:

$12.65

Step-by-step explanation:

i caculated that

5 0

3 years ago

What is the square root of 4?fffffffffffffffffffffffffffffffffffffffffffff

Mumz [18]

Step-by-step explanation:

$\sqrt{a}=b\iff b^2=a\\\\\text{Therefore}\\\\\sqrt4=\pm2\ \text{because}\ (-2)^2=4\ \text{and}\ 2^2=4$

3 0

3 years ago

Read 2 more answers

i need help asap please dont type random anwsers, that will result in it being deleted. GIVING BRAINLIEST ONLY TO CORRECT, INCOR

Veseljchak [2.6K]

Answer:

The area of the rectangle TOUR is 80.00 unit².

Step-by-step explanation:

The area of a rectangle is computed using the formula:

$Area\ of\ a\ Rectangle=length\times width$

Since the dimensions of the rectangle are not provided we can compute the dimensions using the distance formula for two points.

The distance formula using the two point is:

$distance=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Considering the rectangle TOUR the area formula will be:

Area of Rectangle TOUR = TO × OU

The co-ordinates of the four vertices of a triangle are:

T = (-8, 0), O = (4, 4), U = (6, -2) and R = (-6, -6)

Compute the distance between the vertices T and O as:

$TO=\sqrt{(4-(-8))^{2}+(4-0)^{2}}\\=\sqrt{12^{2}+4^{2}} \\=\sqrt{160} \\=4\sqrt{10}$

Compute the distance between the vertices O and U as:

$OU=\sqrt{(6-4)^{2}+(-2-4)^{2}}\\=\sqrt{2^{2}+6^{2}} \\=\sqrt{40} \\=2\sqrt{10}$

Compute the area of rectangle TOUR as follows:

$Area\ of\ TOUR=TO\times OU\\=4\sqrt{10}\times 2\sqrt{10}\\=80\\\approx80.00 unit^{2}$

Thus, the area of the rectangle TOUR is 80.00 unit².

6 0

3 years ago

Can someone help me with this problem

Alex_Xolod [135]

Answer:

Step-by-step explanation:

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

A(-3, 0) , B(3, 2)

$AB=\sqrt{(3-(-3))^2+(2-0)^2}=\sqrt{6^2+2^2}=\sqrt{36+4}=\sqrt{40}=\sqrt{4\cdot10}=\sqrt4\cdot\sqrt{10}=2\sqrt{10}$

A(-3, 0), D(-2, -3)

$AD=\sqrt{(-2-(-3))^2+(-3-0)^2}=\sqrt{1^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}$

AB = CD and AD = BC

The area of a rectangle:

$A=(AB)(AD)$

Substitute:

$A=(2\sqrt{10})(\sqrt{10})=(2)(10)=20$

The perimeter of a rectangle:

$P=2(AB+AD)$

Substitute:

$P=2(2\sqrt{10}+\sqrt{10})=2(3\sqrt{10})=6\sqrt{10}$

4 0

3 years ago