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3 years ago

How many groups of 25 are there in 500

Mathematics

2 answers:

pshichka [43]3 years ago

7 0

If you divide 25 by 500 the answer is 20 therefore there are 20 groups of 25 in 500

Elis [28]3 years ago

6 0

500÷25

they are 20 groups

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Can u explain when answering thanks !

Margarita [4]

I believe its E because you add the fractions together and 12 is alone so yeah... hope this helps

3 0

3 years ago

If sin ⁡x=7/25, and 0 < x < π/2, what is cos⁡(x−3π/2)?

Likurg_2 [28]

Answer:

$Cos(x-\frac{3\pi}{2})=-\frac{7}{25}$

Step-by-step explanation:

Remember that $cos(x-y)=cos(x)cos(y)+sin(x)sin(y)$

Then,

$cos(x-\frac{3\pi}{2})=cos(x)cos(\frac{3\pi}{2})+sin(x)sin(\frac{3\pi}{2})\\$

But $cos(\frac{3\pi}{2})=0$ and $sin(x)=\frac{7}{25}$

Then

$cos(x-\frac{3\pi}{2})=cos(x)*0+\frac{7}{25}*sin(\frac{3\pi}{2})\\=\frac{7}{25}*(-1)=\frac{-7}{25}$

5 0

3 years ago

Two poles are connected by a wire that is also connected to the ground. The first pole is 18 ft tall and the second pole is 24 f

9966 [12]

Answer:

72 feet from the shorter pole

Step-by-step explanation:

The anchor point that minimizes the total wire length is one that divides the distance between the poles in the same proportion as the pole heights. That is, the two created triangles will be similar.

The shorter pole height as a fraction of the total pole height is ...

18/(18+24) = 3/7

so the anchor distance from the shorter pole as a fraction of the total distance between poles will be the same:

d/168 = 3/7

d = 168·(3/7) = 72

The wire should be anchored 72 feet from the 18 ft pole.

_____

<em>Comment on the problem</em>

This is equivalent to asking, "where do I place a mirror on the ground so I can see the top of the other pole by looking in the mirror from the top of one pole?" Such a question is answered by reflecting one pole across the plane of the ground and drawing a straight line from its image location to the top of the other pole. Where the line intersects the plane of the ground is where the mirror (or anchor point) should be placed. The "similar triangle" description above is essentially the same approach.

Alternatively, you can write an equation for the length (L) of the wire as a function of the location of the anchor point:

L = √(18²+x²) + √(24² +(168-x)²)

and then differentiate with respect to x and find the value that makes the derivative zero. That seems much more complicated and error-prone, but it gives the same answer.

7 0

3 years ago

Help me in this question

KATRIN_1 [288]

H = -5t^2 +5t + 10

a) t = 0, h = 10 at time 0

b) h = (-5t + 10) ( t + 1)

c) put 0 in for the height and set each factor = to 0 and solve each
0 = (-5t + 10) and 0 = (t + 1)
solve each t = 2 and t = -1, so t = 2 sec is your solution

d) because a parabola is symmetric, the max will be half way between -1 and 2, at t = 1/2

h = -5(1/2)^2 +5(1/2) + 10
h = -5(1/4) + 5/2 +10
h = -5/4 + 5/2 + 10
h = -5/4 + 10/4 + 40/4
h = 45/4

5 0

3 years ago

How would you model 5 x 2/3?

kenny6666 [7]

That is maybe how and my teacher told me that if you have a whole number you can add a line underneath it

3 0

3 years ago