Answer:
Step-by-step explanation:
⇒49=7×7 =1 × 49
Factors of 49 = 1,7,49
⇒It is given that ,98 is a Multiple of 49.
98 = 49 × 2
So, apart from factors of 49, number 2, and 98 are also included in the set of factors of 98.
→Factors of 98 are =1,2,7,49,98
A method to know the probability is to list down all of the possible combinations that would present an outcome of not more than 5. This is listed below:
1 + 4
4 + 1
2 + 3
3 + 2
1 + 1
2 + 2
1 + 2
2 + 1
1 + 3
3 + 1
There are 10 possible outcomes. But among these, the only ones that can have a sum more than 3 are:
1 + 4
4 + 1
2 + 3
<span>3 + 2
</span>2 <span>+ 2
</span>1 + 3
3 + 1
There are only 7 possible outcomes that meet both requirements. Given that there are 7 outcomes within the total of 10, the probability would be 7/10, or 0.7.
Answer:

Step-by-step explanation:
The lateral surface area of a cone is given by
, where
is the radius of the base of the cone and
is the cone's slant height.
Solving for
:

*Note: It is implied that the surface area of 200 pi square inches is lateral surface area and not total surface area, because it is impossible to obtain a total surface area of 200 pi with a radius of 10 inches.
Primes are
2,3,5,7,11,13...
ok, so it must be multiplule of 2,7 and 11
2*7*11=154
answer is 154
Answer:
Find below the calculations of the two areas, each with two methods. The results are:


Explanation:
<u>A) Method 1</u>
When you are not given the height, but you are given two sides and the included angle between the two sides, you can use this formula:

Where,
is the measure of the included angle.
1. <u>Upper triangle:</u>

2. <u>Lower triangle:</u>

<u></u>
<u>B) Method 2</u>
You can find the height of the triangle using trigonometric properties, and then use the very well known formula:

Use it for both triangles.
3. <u>Upper triangle:</u>
The trigonometric ratio that you can use is:

Notice the height is the opposite leg to the angle of 60º, and the side that measures 100 units is the hypotenuse of that right triangle. Then:


3. <u>Lower triangle:</u>
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<u />
<u />
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