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3 years ago

Help!!! please what's is the figure

Mathematics

2 answers:

uranmaximum [27]3 years ago

8 0

Answer:

Total Area = 108 cm^2

Step-by-step explanation:

Area Trapezoid

Givens

b1 = 6cm

b2 = 12 cm

h = 6cm

Formula

Area = 1/2 (b1 + b2)*h

Solution

Area = 1/2 (6 + 12)*6

Area = 1/2 18*6

Area = 54 cm^2

Triangle

Givens

b = 12 cm

h = 15 - 6

h = 9

Formula

Area = 1/2 * b * h

Solution

Area = 1/2 * 12 * 9

Area = 54 cm^2

Total Area

Total Area = 54 + 54 = 108 cm^2

S_A_V [24]3 years ago

3 0

Hello from MrBillDoesMath!

Answer:

108 cm^2

Discussion:

The area of the logo =

area of Trapezoid on top (I) + area of triangle on bottom (II)

I = (1/2) h ( b1 + b2) h = altitude, b1, b2 = bases

= (1/2) 6 ( 6 + 12)

= (1/2) 6 ( 18)

= 54

II = (1/2) b h

= (1/2) (12) ( 15 - 6) => altitude of triangle = 15 - 6 NOT (1/2) 15

= 54

I + II= 54 + 54 = 108

Thank you,

MrB

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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.

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

Help!!! please what's is the figure​

Help!!! please what's is the figure