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3 years ago

A bicycle tire has a radius of 13.25 inches. How far will the bicycle travel in 40 rotations of the tire?

Mathematics

1 answer:

Effectus [21]3 years ago

8 0

Answer:

it will travel 69 feet on the bike

Step-by-step explanation:

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What is the quotient if you divide 9.12 by 1.14? * 0.08 0.8 8 80

PolarNik [594]

Answer:

Step-by-step explanation:

Divide 9.12 by 1.14

8 0

3 years ago

Read 2 more answers

If you have $45 and they take 2.1% of it how much do you have left ?

IRINA_888 [86]

First you need to make a multiplier for 100-2.1%, to do this you minus the values and divide it by 100:
100-2.1 = 97.9
97.9/100 = 0.979

Then you multiply all the questions by it:
$45 * 0.979 = $44.055
$765 * 0.979 = $748.935
$1300 * 0.979 = $1272.7

Hope this helps! :)

3 0

3 years ago

Write the rule for the myth term -4, 20, -100,

Gekata [30.6K]

You would times it by -5

20/-4 = -5

-100/20 = -5

6 0

3 years ago

FACTOR. THEOREM is x-3 a factor?!

Oduvanchick [21]

Answer:

Yes, (x - 3) is a factor of P(x) and 3 is a zero or root of P(x)

Step-by-step explanation:

Determine whether or not (x - 3) is a factor by using synthetic division with +3 as the divisor. The coefficients of the polynomial p(x) are {2 -5 -4 0 9}.

Setting up synthetic division:

3 / 2 -5 -4 0 9

6 3 -3 -9

-------------------------------

2 1 -1 -3 0

Since the remainder is zero (0), we know that 3 is a root of the polynomial P(x) and that (x - 3) is a factor of said polynomial.

3 0

3 years ago

AB=diameter of the circle O. OC=radius. Arc AXB is an arc of the circle with centre C. Prove areas of the shaded region are =

alexandr1967 [171]

1.
Draw a circle with center C And radius CA, as shown in the attached picture.

Let the lengths of radii AO, OB, OC be R. Triangle ABC is inscribed in the circle with center O and one of its sides is a diameter, this means that the angle ACB is a right angle.
|AO|=|OC|=R, by the Pythagorean theorem |AC|= $\sqrt{2}R$ .

these are all shown in the picture.

2.

Area of triangle ABC is 1/2 * 2R * R= R^2

3.

Let the area between arc BXA and chord AB be Y. (the yellow region).

and let G be the shaded region between arcs AB and AXB.

G=1/2(Area circle with center O)-Y
= $\frac{1}{2} \pi R^{2}-Y$

To find Y:

Notice that the area of the sector ACB is 1/4 of the area of circle with center C, since m(ACB) is 1/4 of 360 degrees.

So Area of sector ACB = $\frac{1}{4} \pi (\sqrt{2} R)^{2}=\frac{1}{4} \pi*2 R^{2} =\frac{1}{2} \pi R^{2}$

Y =area of sector ABC-Area(triangle ABC)= $\frac{1}{2} \pi R^{2}- \frac{1}{2}*2R*R=\frac{1}{2} \pi R^{2}- R^{2}$

4.

Finally,

$G=\frac{1}{2} \pi R^{2}-Y=\frac{1}{2} \pi R^{2}-(\frac{1}{2} \pi R^{2}- R^{2})=R^{2}$

This proves that the 2 shaded regions have equal area.

5 0

3 years ago