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fgiga [73]
3 years ago
13

A 15 ft pole is leaning against a tree. The bottom of the pole is 9 feet away from the bottom of the tree , Approximately how hi

gh up the tree does the top of the pole reach?
Mathematics
1 answer:
tigry1 [53]3 years ago
8 0

Answer:

12ft

Step-by-step explanation:

you do 15squared×9squared to get 144 then you square root it to make 12

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Paul jogged 3.25 miles, rick jogged 3 1/3 miles and sean jogged 3 1/8 miles. list the boys from least to greatest distance jogge
Novosadov [1.4K]

Answer:

Sean jogged 3.125, Paul jogged a distance of 3.25, Rick jogged 3.33

Step-by-step explanation:

make the fractions into decimals once they are in decimal form you cna then go and place them into the correct order.

3 0
3 years ago
G = 15 - m/32
IRINA_888 [86]

The value of 32 in the equation will serve as the time taken by the driver to travel

<h3>Functions and values</h3>

If Alice fills up the gas tank of her car before going for a long drive, and the equation that shows the amount of gas in gallons in Alice's car when she has driven m (miles). is given as:

g = 15 - m/32

The value of 32 in the equation will serve as the time taken by the driver to travel

Learn more on distance and time here; brainly.com/question/17273444

#SPJ1

6 0
2 years ago
Plz answer both if you can
Harlamova29_29 [7]
I hope this helps you



130+x-5+x-35+x+30+75=360


3x=360-195


3x=165


x=55


m (B)=55-5=50
7 0
3 years ago
You own an accessories store, and sales last month were $24,000. You had $3,500 in discounts and $975 in returns.
Len [333]

Answer:

The net sales for last month were <u>$19,525</u>.

Step-by-step explanation:

Given:

Last month sales were $24,000.

Discounts is $3,500 and $975 in returns.

Now, to get the net sales for last month.

So, we deduct the discount:

<em>Sales - discounts</em> = \$24,000 - \$3,500 =\$20,500.

Then, we deduct the returns from the remaining amount:

<em>Sales after discounts - returns</em> = \$20,500 - \$975

                                                   = \$19,525.

Therefore, the net sales for last month were $19,525.

7 0
3 years ago
g Let G be a not necessarily abelian group with normal subgroups H and K such that H contains K (i.e., K ✂ G, H ✂ G, K ≤ H) and
allsm [11]

Answer:

Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, [ab]_K is equal to [ba]_K, thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.

Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, [ab]_H = [ba]_H . Since a and b were generic elements of H, then H/G is abelian.

4 0
3 years ago
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