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Jet001 [13]
3 years ago
8

Please help! This is the last question for my math test and I really need a good grade!

Mathematics
1 answer:
Sphinxa [80]3 years ago
3 0

Answer:

GI

Step-by-step explanation:

this is the correct answer

You might be interested in
If 7 - 1 is 6 then what would it be if you multiply it by 11700 and subtract by 5000?
steposvetlana [31]

Answer: 65200

7 - 1 = 6 x 11700 = 70200 - 5000 = 65200

8 0
2 years ago
The length of a rectangular garden ABCD is 9 feet more than its width. It is surrounded by a brick walkway 4 feet wide as shown
Anna71 [15]

Answer:

  16.5 ft by 25.5 ft

Step-by-step explanation:

Let w represent the width of the garden in feet. Then w+9 is the garden's length, and w(w+9) represents its area.

The surrounding walkway adds 8 feet to each dimension, so the total area of the garden with the walkway is ...

  (w+8)(w+9+8) = w^2 +25w +136

If we subtract the area of the garden itself, then the remaining area is that of the walkway:

  (w^2 +25w +136) - (w(w+9)) = 400

  16w + 136 = 400 . . .simplify

  16w = 264 . . . . . . . . subtract 136

  264/16 = w = 16.5 . . . . . width of the garden in feet

  w+9 = 25.5 . . . . . . . . . . .length of the garden in feet

5 0
4 years ago
Nickel-63 has half-life of about 96 years. after 336 years,about how many milligrams of an 800 mg sample will remain?
vagabundo [1.1K]
F=ar^t if the half-life is 96 years...

.5=r^96

ln(.5)=96lnr

ln(.5)/96=lnr

r=e^((ln.5)/96)

f=800e^(336(ln.5)/96)

f=70.71

So about 71mg will remain after 336 years.


5 0
3 years ago
Read 2 more answers
20+ POINTS!!!!
azamat
A, b, d.  Dilation is NOT <span>an </span><span>isometry</span>
6 0
3 years ago
Find how many solutions there are to the given equation that satisfy the given condition.
den301095 [7]

Answer:

17550 solutions

Step-by-step explanation:

Given that:

y1 +y2+y3+y4=27

where;

(yi  ≥ 0 and yi \epsilon {\displaystyle \mathbb {Z} }  )

The no. of a nonnegative integer determines the number of ways to choose 27 objects from (4) distinct objects with repetition regardless of the order.

i.e

\bigg(^{27}_{4} \bigg)

∴

The number of nonnegative integer solution is \bigg(^{27}_{4} \bigg)

= \dfrac{27!}{4!(27-4)!}

= \dfrac{27!}{4!(23)!}

= \dfrac{27\times 26\times 25\times 24 \times  23 !}{4\times 3\times 2\times 1(23)!}

= \dfrac{27\times 26\times 25\times 24 }{4\times 3\times 2\times 1}

= \dfrac{421200}{24}

= 17550 solutions

7 0
3 years ago
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