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uysha [10]
3 years ago
11

How do you write 2.59 x 10-3 in standard form?

Mathematics
1 answer:
Allisa [31]3 years ago
7 0

Answer:

2.59 x 10-3 = 22.9

and 22.9 in standard form is Your number written in expanded form is:

20 + 2 + 0.9

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3 years ago
For the following rectangular prism, find the length of the diagonal connecting A to B. Round to the nearest tenth.
meriva

Answer:

I think it is 192

Step-by-step explanation:

Volume of cube: 5 x 4 x 8 = 160

Area of diagonal: 160/2 = 80

Length of diagonal: 80 / 0.5 x 3 = 480

AB = 480 / 5 x 2 = 192

8 0
2 years ago
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PIT_PIT [208]

Answer:

7/6

Step-by-step explanation:

\frac{ \frac{x}{4}  +  \frac{x}{3} }{ \frac{x}{2} }  = \frac{x( \frac{1}{4}  +  \frac{1}{3} )}{ x(\frac{1}{2}) } \\  =  \frac{ \frac{1}{4}  +  \frac{1}{3} }{ \frac{1}{2} }  \\  =  \frac{14}{12}  =  \frac{7}{6}

3 0
2 years ago
Read 2 more answers
△PQR has two known interior angles of 34°, and 80°. △RST has two known interior angles of 24°, and 90°. What can be determined a
Nonamiya [84]
B.the triangles are not similar
7 0
3 years ago
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dexar [7]

Answer: Anything between 0 and 10, excluding both endpoints.

In terms of symbols we can say 0 < w < 10 where w is the width.

===================================================

Explanation:

You could do this with two variables, but I think it's easier to instead use one variable only. This is because the length is dependent on what you pick for the width.

w = width

2w = twice the width

2w-5 = five less than twice the width = length

So,

  • width = w
  • length = 2w-5

which lead to

area = length*width

area = (2w-5)*w

area = 2w^2-5w

area < 150

2w^2 - 5w < 150

2w^2 - 5w - 150 < 0

To solve this inequality, we will solve the equation 2w^2-5w-150 = 0

Use the quadratic formula. Plug in a = 2, b = -5, c = -150

w = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\w = \frac{-(-5)\pm\sqrt{(-5)^2-4(2)(-150)}}{2(2)}\\\\w = \frac{5\pm\sqrt{1225}}{4}\\\\w = \frac{5\pm35}{4}\\\\w = \frac{5+35}{4} \ \text{ or } \ w = \frac{5-35}{4}\\\\w = \frac{40}{4} \ \text{ or } \ w = \frac{-30}{4}\\\\w = 10 \ \text{ or } \ w = -7.5\\\\

Ignore the negative solution as it makes no sense to have a negative width.

The only practical root is w = 10.

If w = 10 feet, then the area = 2w^2-5w results in 150 square feet.

----------------------

Based on that root, we need to try a sample value that is to the left of it.

Let's say we try w = 5.

2w^2 - 5w < 150

2*5^2 - 5*5 < 150

25 < 150 ... which is true

This shows that if 0 < w < 10, then 2w^2-5w < 150 is true.

Now try something to the right of 10. I'll pick w = 15

2w^2 - 5w < 150

2*15^2 - 5*15 < 150

375 < 150 ... which is false

It means w > 10 leads to 2w^2-5w < 150  being false.

Therefore w > 10 isn't allowed if we want 2w^2-5w < 150 to be true.

4 0
2 years ago
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