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3 years ago

A semicircle is attached to the side of a rectangle as shown.

Mathematics

1 answer:

Genrish500 [490]3 years ago

6 0

Answer:

$58.8\ mm^{2}$

Step-by-step explanation:

we know that

The area of the figure is equal to the area of a rectangle plus the area of semicircle

Step 1

Find the area of the rectangle

The area of the rectangle is equal to

$A=bh$

we have

$b=9\ mm$

$h=3\ mm$

substitute

$A=9*3=27\ mm^{2}$

Step 2

Find the area of semicircle

The area of semicircle is equal to

$A=\frac{1}{2}\pi r^{2}$

we have

$r=9/2=4.5\ m$

substitute

$A=\frac{1}{2}(3.14)(4.5^{2})=31.8\ mm^{2}$

Step 3

Find the area of the figure

$27\ mm^{2}+31.8\ mm^{2}=58.8\ mm^{2}$

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Karla invests $300 compounded every 6 months at a rate of 10% for 3 years. At the end of three years, Karla will have $402.03 in

adell [148]

Answer:

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1 + r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = $300

r = 10% = 10/100 = 0.1

n = 2 because it was compounded 2 times in a year(6 months).

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Therefore,

A = 300(1 + 0.1/2)^2 × 3

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A = 300(1.05)^6

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4 0

3 years ago

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KonstantinChe [14]

Answer:

21 h^10

Step-by-step explanation:

(7h^3)(3h^7)

Lets multiply

7*3 h^3 h^7

When the bases are the same , we add the exponents (x^a * x^b) = x ^ (a+b)

21 h^(3+7)

21 h^10

5 0

3 years ago

Read 2 more answers

Can someone please help me

Aleks04 [339]

The answer is D. 60/100

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3 years ago

What is the constant in the expression 9x +5y + 12

Ber [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

evaluate the line integral ∫cf⋅dr, where f(x,y,z)=5xi−yj+zk and c is given by the vector function r(t)=⟨sint,cost,t⟩, 0≤t≤3π/2.

meriva

We have

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} \vec f(\vec r(t)) \cdot \dfrac{d\vec r}{dt} \, dt$

and

$\vec f(\vec r(t)) = 5\sin(t) \, \vec\imath - \cos(t) \, \vec\jmath + t \, \vec k$

$\vec r(t) = \sin(t)\,\vec\imath + \cos(t)\,\vec\jmath + t\,\vec k \implies \dfrac{d\vec r}{dt} = \cos(t) \, \vec\imath - \sin(t) \, \vec\jmath + \vec k$

so the line integral is equilvalent to

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (5\sin(t) \cos(t) + \sin(t)\cos(t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (6\sin(t) \cos(t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (3\sin(2t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \left(-\frac32 \cos(2t) + \frac12 t^2\right) \bigg_0^{\frac{3\pi}2}$

$\displaystyle \int_C \vec f \cdot d\vec r = \left(\frac32 + \frac{9\pi^2}8\right) - \left(-\frac32\right) = \boxed{3 + \frac{9\pi^2}8}$

7 0

2 years ago