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3 years ago

A triangle has a height that is 4 /3 the length of its base. If the length of its base is 24 cm, what is its area?

2 answers:

6 0

Triangle Area = ½ * height * base

base = 24 therefore
height = (4/3) * base = (4/3) * base = 32 cm

Triangle Area = ½ * 32 * 24
= 384 cubic centimeters

elena55 [62]3 years ago

5 0

Answer:

384

Step-by-step explanation:

on usatestprep it is b

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An 8 foot square floor is to be covered with square tiles measuring 8 inches on each side. If each tile costs 50 cents, how much

wariber [46]

9514 1404 393

Answer:

Step-by-step explanation:

Each tile is 8/12 ft = 2/3 ft on a side. Then 8/(2/3) = 12 tiles will fit along each edge of the square area to be tiled. That is ...

12 × 12 = 144

tiles will be needed to cover the area.

The cost of 144 tiles at $0.50 each is ...

(144)($0.50) = $72.00

8 0

3 years ago

A scale drawn on the map shows that 1 inch represents 20 miles. If two cities

spin [16.1K]

Answer:

50 miles

Step-by-step explanation:

20 multiplied by 2.5 = 50

8 0

2 years ago

Let ρ = x3 + xe−x for x ∈ (0, 1), compute the center of mass.

hram777 [196]

The center of mass is mathematically given as

$\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

<h3>What is the center of mass.?</h3>

Determine the center of mass in one dimension:

Represent the masses at the respective distances.

$\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \ \ \ x \in(0,1)$ \\\end$

We calculate the total mass of the system.

$\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}$

Step 03: Calculate the moment of the system.

$\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}$

we calculate the center of mass.

$\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}$

Read more about the center of mass.

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8 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B28%7D%7B40%7D" id="TexFormula1" title="\frac{28}{40}" alt="\frac{28}{40}" align="abs

Vitek1552 [10]

Answer:

7/10

Step-by-step explanation:

28/40=

14/20 =

7/10

8 0

3 years ago

Read 2 more answers

Please help me with this question!!!!!

amm1812

Answer:

B. $H(x)=-.008(x-82)^{2}+83$

Step-by-step explanation:

Well lets graph the following equation first,

H(x) = -.011(x - 82)^2 + 75

Then lets graph all the other equations.

Look at the image below↓

So A.

The purple line does not go farther or higher,

so A. is incorrect

The black line goes higher and farther,

meaning it is correct.

The red line goes farther but not higher.

So it is incorrect

Again the blue line does not go farther of higher.

So its wrong.

<em>answer choice B. </em> $H(x)=-.008(x-82)^{2}+83$ <em>is correct.</em>

<em />

<em>Hope this helps :)</em>

3 0

3 years ago