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ryzh [129]
1 year ago
15

Solve each of the following equation

Mathematics
1 answer:
natali 33 [55]1 year ago
7 0
(9)^{2x}(27)^{1-x}=(81)^{2x+1}

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Need help please!<br><br> (5^3)^6
Umnica [9.8K]

Answer:

5^{18}=3,814,697,265,625

Step-by-step explanation:

Multiply the exponents

3×6=18

5^{18}

3 0
3 years ago
Read 2 more answers
Which of the following are true statements.
Mariulka [41]

Answer:

Second statement is true.

The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

Step-by-step explanation:

for first part of statement

The lengths 7, 40 and 41 can not be sides of a right triangle.

If the square of long side is equal to the sum of square of other two sides

then the given length can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

c^{2} =a^{2} +b^{2}----------(1)

Let c=41 and a = 7 and b=40

Put all the value in equation 1.

41^{2} =7^{2} +40^{2}

1681=49+1600

1681=1649

Therefore, the square of long side is not equal to the sum of square of other two sides, So given lengths 7, 40 and 41 can not be sides of a right triangle.

for second part of statement.

The lengths 12, 16, and 20 can be sides of a right triangle.

Check the given length by Pythagoras Theorem.

Let c=20 and a = 12 and b=16

20^{2} =12^{2} +16^{2}

400=144+256

400=400

Therefore, the square of long side is equal to the sum of square of other two sides, So given the lengths 12, 16, and 20 can be sides of a right triangle.

Therefore, The lengths 7, 40 and 41 can not be sides of a right triangle. The lengths 12, 16, and 20 can be sides of a right triangle.

8 0
3 years ago
Kyleigh is making a bowl of punch. The recipe calls for 2 1/4 cups of sherbet. She uses 1 2/3 of that amount. How much sherbet d
Delvig [45]

Answer:

I don't know if this is absolutely correct but i got 1.35 by dividing 1 2/3 by 2 1/4

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Find the slope of the line that passes through the pair of points. (3,-3), (3,-4)
Ivan
The slope of this line is undefined as it is a vertical line.

Hope this helps.
5 0
3 years ago
Given cscx =-5/2 between 270 and 360 find sin2x
ICE Princess25 [194]
Hello : 
cscx = -5/2
1/sinx =-5/2
sinx=-2/5
sin²x+cos²x=1
4/25+cos²x=25/25
cos²x = 21/25
between 270 and 360 : cosx > 0          cosx=(<span>√21)/5
but : sin2x =2sinx cosx
       sin2x = 2(-2/5)(</span>(√21)/5)
       sin2x = (-4√21)/25
3 0
3 years ago
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