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icang [17]
4 years ago
10

Help with number 6. Plsss

Mathematics
1 answer:
Dmitriy789 [7]4 years ago
6 0

Answer:

3×10^-1

Step-by-step explanation:

use standard form since an integer is a whole number

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A silo is constructed using a cylinder with a hemisphere on top. The circumference of the hemisphere and the circumference of th
givi [52]
1. The surface that will be exposed to the rain is the lateral surface of the cylinder and the hemisphere.

2. The lateral area of the cylinder is a rectangle with height equal to the height of the cylinder and width equal to the circumference of the base of the cylinder.

3. the circumference of the base = 2\pir=2*\pi*5=10\pi
so the lateral area = 10\pi*h=10\pi*40=400\pi (ft^{2})

4. The surface area of a sphere is 4 \pi  r^{2}
so the surface of the hemisphere = \frac{1}{2} *4 \pi r^{2}=2 \pi (5)^{2}=50 \pi(ft^{2})

5. Total area exposed = 450\pi ft^{2}
4 0
4 years ago
Read 2 more answers
Can someone please help me with b, d, and 9-13? I already drew a number line, thanks in advance! :)
inessss [21]
B)
\frac{3}{13}  = s
d)
\frac{5}{4}  = s
9) I'm not really sure how to answer it, but I guess like 2.1, 2.2, 2.5, etc.,

10)
{2}^{2}  = 4
{3}^{2}  = 9
so {5, 6, 7, 8} is the answer

11) for some reason it won't let me insert a picture but put place
\sqrt{4}
above 2, and then place
\sqrt{9}
above 3, and then place
\sqrt{7}
in between 2 and 3, but place it a little closer to three since
\sqrt{7}  = 2.65
12) place
\sqrt{5}
between 2 and three, but closer to two since
\sqrt{5}  = 2.24
and also
\sqrt{5}  <  \sqrt{7}
and for number 13) the square root of 144 is 12, and the square root of 169 is 13, so any numbers between 12 and 13 will work.


I hope this helped and if not, message me and ill try to explain!
8 0
3 years ago
Is 1 1/4 a rational number?
lisabon 2012 [21]

Answer:

Hey there!

1 1/4 is a rational number because it can be expressed as a fraction.

Let me know if this helps :)

3 0
3 years ago
We put 200 balls into 100 boxes such that every box got at least 1 ball and at most 100 balls. Prove that there are some boxes t
Vinvika [58]

Explanation:

Lets first show that al least 50 of the boxes contail at most 2 balls.

If there were 50 + k boxes with 3 balls or more, then we should have 100 - (50 + k) = 50 - k balls with 1 ball or 2. However in those 50 + k boxes with 3 or more balls we have alredy at least 3*(50+k) = 150 + 3k balls in them, and the amount of balls remaining is, as a result, at most 200 - (150 + 3k) = 50 - 3k, which cant be fit in 50 - k balls if we put at least 1 on each.

Therefore, there are at least 50 boxes with 1 or 2 balls. Whithin those boxes, we can obtain any number of balls selecting the appropiate boxes. Lets assume that we want M balls, and we have A boxes with 2 balls and B boxes with 1 ball, we have this possibilities (M equal or less than 2A + B, the total number of balls):

  • If M > 2A, then we pick all boxes with 2 balls (A in total) and M - 2A boxes with 1 ball. We have 2*A + (M-2A) = M. We are able to pick M - 2A boxes because B ≥ M - 2A.
  • If M ≤ 2A, and it is even, then we pick M/2 boxes with 2 balls.
  • If M ≤ 2A and it is odd, then we pick (M-1)/2 boxes with 2 balls and 1 box with 1 ball (if all boxes contain 2 balls or more, then we could pick 50 boxes with 2 balls because at least 50 boxes contain 1 or 2 balls; so we can assume that at least 1 box contain one single ball).

Lets call C the sum of the balls in the boxes with 1 or 2 balls. C should be at least 50. The argument made previously shows that we can pick boxes of 1 or 2 balls that cover any number of balls below to C. This means that we can obtain any number below 50; furthermore, if C is equal or greater than 100, then the problem is alredy solved. Lets suppose that C is lower than 100.  This means that the other boxes contain more than 100 balls in total.

Since we cant put more than 100 balls in one single box, then there should be a combination of boxes with 3 or more balls that contain between 50 and 100 balls. If that is not the case, then lets call L the biggest number of balls below 50 that we can obtain with boxes with 3 or more balls. Since the sum of all balls is bigger than 100, then there should be a box outside those we use to obtain L with 3 or more balls. Since L was the biggest number we could obtain below 50, and we are supposing that we cant obtain any number between 50 and 100, then that box should have more than 50 balls. Which means that that box alone could be used to obtain a number between 50 and 100. This is a contradiction.

The paragraph above shows that we can make a combination of boxes with 3 or more balls which combined number of balls is a number N between 50 and 100. Since we can make any number between 0 and 50 with boxes, for example 100 - N, with boxes of 1 or 2 balls, then we should be able to make exactly 100 balls using the boxes we have available.

I hope that works for you!

4 0
3 years ago
Someone, please help me with this.
Alex Ar [27]

Answer:

\large\boxed{x=14}

Step-by-step explanation:

Because the angles are adjacent to each other and form a straight line, they are supplementary angles. Supplementary angles add up to 180 degrees, therefore, you can add the two angle measures together and set them equal to 180.

(7x - 1) + (6x - 1) = 180 combine like terms

13x - 2 = 180 add 2 to both sides

13x = 182 divide by 13 on both sides

\boxed{x=14}

5 0
4 years ago
Read 2 more answers
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