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3 years ago

5 time 5 to the power of 2 simplified

Mathematics

2 answers:

Vikki [24]3 years ago

4 0

The answer is 20 fasho

Maksim231197 [3]3 years ago

3 0

Im not sure wym but i think the answer is 20

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Answer questions 15 and 17 show work pls

fredd [130]

Answer:

#15) B. 30 mn^5

#17) B. 1/2

Step-by-step explanation:

The area of a trapezoid is given in the formula: 1/2(a + b) * h, where a is the length of the top of the trapezoid, b is the length of the bottom of the trapezoid, and h is the height of the trapezoid.

All of these measurements are given so all that you need to do is to substitute these values into the formula.

Substitute 3 for a, 9 for b, and 5 for h.

1/2(3 + 9) * 5

Solve inside the parentheses first. Add 3 and 9.

1/2(12) * 5

Multiply 12 and 1/2 together.

6 * 5

Multiply 6 and 5.

We need to figure out if the area is to the 5th or 6th power. When we added 3 and 9 together, we combined like terms so the exponent stayed to the 3rd power.

After multiplying this ^3 by the 5mn^2, the exponent becomes to the 5th power because you add exponents when multiplying.

Therefore the final answer is B. 30 mn^5.

When going down from 32 to 8 to 2, you can see that each number is being divided by 4.

32 / 4 = 8...

8 / 4 = 2...

So to find the next number in this sequence you would divide 2 by 4.

2 / 4 = 1/2

The answer is B. 1/2.

6 0

3 years ago

A man travelled 3/5 of his journey by rail, 1/4 by a taxi, 1/8 by a bus and the remaining 2 km on foot.What is the length of his

docker41 [41]

Hello friend,here is the solution

<span>Let x be the total distance </span>

<span>Distance travelled by rail = 3/5 x </span>

Distance travelled by taxi = 1/4 x

<span>Distance travelled by bus = 1/8 x </span>

<span>The equation we obtain is </span>
========================

<span>x = 3/5 x + 1/4 x + 1/8 x +2 </span>

______________________________________

<span>hope this solution helps you......</span>

7 0

3 years ago

Three moons are in the same circular orbit around a planet. The moons are each 120,000 kilometers from the surface of the planet

VladimirAG [237]

Answer:

Step-by-step explanation:

Alright, lets get started.

Please refer the diagram I have attached.

MN is the diameter of the planet which is $MN = 60000$

AM and NC are the distance of moons from the surface of the planet.

$AM = NC = 120000$

Since, angle ABC is given as 90 degree, it means, line AC will pass from the diameter of the planet.

So, distance between moon A and moon C is =

$AC=AM + MN + NC$

$AC = 120000+ 60000+ 120000$

$AC = 300000$

Hence the distance between point A and point C is 300000 Km. : Answer

Hope it will help :)

8 0

3 years ago

Write an expression that equals 56 include a n exponent

Ket [755]

56^1 is a easy way to do a exponent I think

3 0

3 years ago

Pls help if pls do most important 15 number 20 number 26 number 28 number 30 number 17 number 25 number 22 number 24 number if u

Tamiku [17]

Answer:

Step-by-step explanation:

15. $\frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx} \\$

16. $\frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }$

$=> (1+cos\alpha )(1-cos\alpha ) = sin^{2} \alpha \\=> 1-cos^{2}\alpha =sin^{2} \alpha \\=> sin^{2}\alpha +cos^{2}\alpha =1\\$

=> The clause is correct

17. Do the same (16)

18. $\frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx$

19.

$\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\$

=> The clause is correct

20. Do the same (18)

21. Left = $\frac{1}{cosecA+cotA} = \frac{1}{\frac{1}{sinA}+\frac{cosA}{sinA} } = \frac{1}{\frac{1+cosA}{sinA} }= \frac{sinA}{1+cosA}$

Right = $cosecA-cotA = \frac{1}{sinA}-\frac{cosA}{sinA}= \frac{1-cosA}{sinA} \\$

Same with (16) => Left = Right => The clause is correct

22. Do the same (21)

Too long, i'm so lazy :))))

5 0

3 years ago