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1 year ago

The circumference of the hub cap of a tire is 80.07 centimeters. find the area of this hub cap. use

Mathematics

1 answer:

NNADVOKAT [17]1 year ago

6 0

The area of this hub cap is about 510.44625 cm².

<h3>What is Area of Circle?</h3>

The area of circle is pi times the square of the radius.

Area of circle =π r²

Circumference (C) = 2π r

80.07 = 2 (3.14) r

76.87/ (2x 3.14) =r

r = 12.75 cm

now, Area be

=π r²

= 3.14*12.5*12.75

=510.44625 cm²

Learn more about this concept here:

brainly.com/question/16866711

#SPJ1

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The Eiffel Tower is about 301 meters tall. Cooper is making a scale model of it using the scale 750 meters : 1 meter.

Leto [7]

301/750=

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2 years ago

The quotient of (x^3+3x^2-4x-12)/(x^2+5x+6)

Elis [28]

Answer: The quotient is (x-2).

Step-by-step explanation:

Since we have given that

$f(x)=(x^3+3x^2-4x-12)\\\\and\\\\g(x)=x^2+5x+6\\\\So,\ \frac{\left(x^3+3x^2-4x-12\right)}{\left(x^2+5x+6\right)}$

Now, we have to find the quotient of the above expression.

So, here we go:

$Factorise\ (x^3+3x^2-4x-12)\\\\=\left(x^3+3x^2\right)+\left(-4x-12\right)\\\\=-4\left(x+3\right)+x^2\left(x+3\right)\\\\=\left(x+3\right)\left(x^2-4\right)$

Now, we will divide the above simplest form with g(x):

$\frac{\left(x+3\right)\left(x^2-4\right)}{\left(x+2\right)\left(x+3\right)}\\\\=\frac{x^2-4}{x+2}\\\\=\frac{\left(x+2\right)\left(x-2\right)}{x+2}\ using\ (a^2-b^2)=(a+b)(a-b)\\\\=x-2$

Hence, the quotient is (x-2).

6 0

2 years ago

Read 2 more answers

What are the amplitude, period, and phase shift of the given function? f(t)=-1/2sin(3t-2pi)

Romashka [77]

Answer:

The $amplitude=|A|=\frac{1}{2}$ , $period=3$ and $\text{phase shift}=\frac{2\pi}{3}$ .

Step-by-step explanation:

The given function is

$f(t)=-\frac{1}{2}\sin (3t-2\pi)$ .... (1)

The general form of sine function is

$F(t)=A\sin (Bt-C)+D$ ....(2)

where, |A| is the amplitude, B is period, D is the vertical shift (up or down), and C/B is used to find the phase shift.

On comparing (1) and (2), we get

$A=-\frac{1}{2}$

$B=3$

$C=2\pi$

$D=0$

So,

$amplitude=|A|=\frac{1}{2}$

$period=3$

$\text{phase shift}=\frac{2\pi}{3}$

Therefore $amplitude=|A|=\frac{1}{2}$ , $period=3$ and $\text{phase shift}=\frac{2\pi}{3}$ .

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3 years ago

Given that lines q and m are parallel and are cut by two

sashaice [31]

angle 1 = 180-110 = 70 degrees (linnear pair)

angle 2= 180-115 = 65 degrees (linnear pair)

angle 3 = angle 2 = 65 degrees (corresponding angles)

angle 4= 180 - angle 3 = 180-65 = 115 degrees (linnear pair)

(your teacher might not like this, depends on the cituation, but use 180 degrees in triangle to get 7)

angle 7 = 180 -angle1 -angle 2= 180-70-65= 45 degrees

angle 6 = angle 7 = 45 degrees ( vertical angles)

angle 8= 180- angle 7= 180-45 = 135 ( linnear pair)

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3 years ago

PLEASE HELP ME IM STRUGGLING!!!

Kitty [74]

Answer:

The required answer is $c=7\sqrt{3}$

Therefore the number in green box should be 7.

Step-by-step explanation:

Given:

AB = 7√2

AD = a , BD = b , DC = c , AC = d

∠B = 45°, ∠C = 30°

To Find:

c = ?

Solution:

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$\sin B = \dfrac{\textrm{side opposite to angle B}}{Hypotenuse}\\$

Substituting the values we get

$\sin 45 = \dfrac{AD}{AB}= \dfrac{a}{7\sqrt{2}}$

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Now in Triangle ADC Tangent identity we have

$\tan C = \dfrac{\textrm{side opposite to angle C}}{\textrm{side adjacent to angle C}}$

Substituting the values we get

$\tan 30 = \dfrac{AD}{DC}= \dfrac{a}{c}\\\\\dfrac{1}{\sqrt{3}}=\dfrac{7}{c}\\\\\therefore c=7\sqrt{3}$

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3 years ago