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3 years ago

What are all of the factors for 100 and 73. Please help with them and write them separate. thanks.

Mathematics

1 answer:

Zolol [24]3 years ago

6 0

The factors of 100 are 1, 2, 4, 5, 10, 20, 25 50 and 100
the factors of 73 are 1 and 73. 73 os a prime number

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3 is 12% of what number

AleksAgata [21]

3/12% = 25

<em>Therefore, 3 is 12% of 25.</em>

3 0

3 years ago

Read 2 more answers

Suppose a circle has a radius of 13 inches. How far would a 24 inch chord be from the center of the circle? (Hint: Draw a diagra

lys-0071 [83]

Answer:

Step-by-step explanation:

Directions

Draw a circle
Dear a chord with a length of 24 inside the circle. You just have to label it as 24
Draw a radius that is perpendicular and a bisector through the chord
Draw a radius that is from the center of the circle to one end of the chord.
Label where the perpendicular radius to the chord intersect. Call it E.
You should get something that looks like the diagram below. The only thing you have to do is put in the point E which is the midpoint of CB.

Givens

AC = 13 inches Given

CB = 24 inches Given

CE = 12 inches Construction and property of a midpoint.

So what we have now is a right triangle (ACE) with the right angle at E.

What we seek is AE

Formula

AC^2 = CE^2 + AE^2

13^2 = 12^2 + AE^2

169 = 144 + AE^2 Subtract 144 from both sides.

169 - 144 = 144-144 + AE^2 Combine

25 = AE^2 Take the square root of both sides

√25 = √AE^2

5 = AE

Answer

The 24 inch chord is 5 inches from the center of the circle.

4 0

2 years ago

Read 2 more answers

Use the distributive property to demonstrate that the product of 10 and -1/5 is a negative number. Make sure to include all step

rodikova [14]

-1(10 * (-1/5))

(-10)(1/5)

-2

4 0

3 years ago

4.) What is the exact value of sinθ when θ lies in Quadrant II and cosθ=−513

dimaraw [331]

Answer:

Part 4) $sin(\theta)=\frac{12}{13}$

Part 10) The angle of elevation is $40.36\°$

Part 11) The angle of depression is $78.61\°$

Part 12) $arcsin(0.5)=30\°$ or $arcsin(0.5)=150\°$

Part 13) $-45\°$ or $225\°$

Step-by-step explanation:

Part 4) we have that

$cos(\theta)=-\frac{5}{13}$

The angle theta lies in Quadrant II

The sine of angle theta is positive

Remember that

$sin^{2}(\theta)+ cos^{2}(\theta)=1$

substitute the given value

$sin^{2}(\theta)+(-\frac{5}{13})^{2}=1$

$sin^{2}(\theta)+(\frac{25}{169})=1$

$sin^{2}(\theta)=1-(\frac{25}{169})$

$sin^{2}(\theta)=(\frac{144}{169})$

$sin(\theta)=\frac{12}{13}$

Part 10)

Let

$\theta$ ----> angle of elevation

we know that

$tan(\theta)=\frac{85}{100}$ ----> opposite side angle theta divided by adjacent side angle theta

$\theta=arctan(\frac{85}{100})=40.36\°$

Part 11)

Let

$\theta$ ----> angle of depression

we know that

$sin(\theta)=\frac{5,389-2,405}{3,044}$ ----> opposite side angle theta divided by hypotenuse

$sin(\theta)=\frac{2,984}{3,044}$

$\theta=arcsin(\frac{2,984}{3,044})=78.61\°$

Part 12) What is the exact value of arcsin(0.5)?

Remember that

$sin(30\°)=0.5$

therefore

$arcsin(0.5)$ -----> has two solutions

$arcsin(0.5)=30\°$ ----> I Quadrant

$arcsin(0.5)=180\°-30\°=150\°$ ----> II Quadrant

Part 13) What is the exact value of $arcsin(-\frac{\sqrt{2}}{2})$

The sine is negative

The angle lies in Quadrant III or Quadrant IV

Remember that

$sin(45\°)=\frac{\sqrt{2}}{2}$

therefore

$arcsin(-\frac{\sqrt{2}}{2})$ ----> has two solutions

$arcsin(-\frac{\sqrt{2}}{2})=-45\°$ ----> IV Quadrant

$arcsin(-\frac{\sqrt{2}}{2})=180\°+45\°=225\°$ ----> III Quadrant

5 0

3 years ago

A sum of rs 400 is lent for 3 yrs at the rate of 6% per annum. Find the interest.

Nimfa-mama [501]

Interest is equal to rs 72

Simple Interest is computed by multiplying the principal to the interest rate and the time.

I = P * r * t

Principal = rs 400
Interest rate = 6%
Term / time = 3 yrs

I = 400 * 6% * 3
I = 72 the total interest that must be paid in 3 yrs.

3 0

3 years ago