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wel
3 years ago
10

If we were in the core of the galaxy...

Mathematics
2 answers:
Svetllana [295]3 years ago
7 0

Answer:

B

Step-by-step explanation:

Mashutka [201]3 years ago
6 0
Second and last option
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Help please!!! | DON'T GUESS, please show work
Aleksandr-060686 [28]
For one of the triangular sides, 2.6 x 4 = 10.4 divided by 2 = 5.2
Other triangle; 2.6 x 3 = 7.8 /2 = 3.9
Rectangular side in front; 6.5 x 3 = 19.5
Back; 6.5 x 4 = 26
Base; 6.5 x 2.6 = 16.9
Answer: For the surface area it is 74.6 (SUPER SORRY IF THIS ISNT RIGHT OMG)
5 0
4 years ago
What is the value of hin the figure below? In this diagram, ABAD ~ ACBD.
VashaNatasha [74]

Answer:

  F.  8

Step-by-step explanation:

The ratio of the long side to the short side is the same in similar triangles. The long side of triangle BAD is AD, which has length 20-4 = 16.

  BD/DE = AD/BD

  h/4 = 16/h

  h^2 = 64 . . . . . . . multiply by 4h

  h = 8 . . . . . . . . . . take the square root (matches selection F)

_____

<em>Comment on this geometry</em>

BD = √(AD·DC) is called the "geometric mean" of the segments AD and DC. This geometry has some other geometric mean relationships as well:

BC = √(AC·DC)

BA = √(AC·AD)

8 0
3 years ago
Luke bought 256256256 ounces of strawberries. He divided them evenly between 888 different pies. How many pounds of strawberries
SIZIF [17.4K]

Answer:

2 pounds.

Step-by-step explanation:

We have been given that Luke bought 256 ounces of strawberries. He divided them evenly between 8 different pies. We are asked to find the pounds of strawberries that Luke put on each pie.

First of all, we will convert 256 ounces into pounds.

We know that 1 pound equals 16 ounces. To convert 256 ounces into pounds, we need to divide 256 by 16.

\text{256 ounces}=\frac{256}{16}\text{ pounds}=16\text{ pounds}

Now we will divide 16 pounds by 8 to find number of strawberries put on each pie.

\text{Pounds of strawberries on each pie}=\frac{16}{8}

\text{Pounds of strawberries on each pie}=2

Therefore, Luke put 2 pounds strawberries on each pie.

4 0
3 years ago
The sum of two positive integers, a and b, is at least 30. The difference of the two integers is at least 10. If b is th
Dafna11 [192]

Answer:

a + b \geq 30

b - a \geq 10

Step-by-step explanation:

Given

The sum of the two positive integer a and b is at least 30, this means the sum of the two positive integer is 30 or greater than 30, so we write the inequalities as below.

a + b \geq 30

The difference of the two integers is at least 10, if b is the greater integer then we subtract integer a from integer b, so we write the inequality as below.

b - a \geq 10

Therefore, the following system of inequalities could represent the values of two positive integers a and b.

a + b \geq 30

b - a \geq 10

6 0
4 years ago
Solve the following equation by completing the square. 3x^2-3x-5=13
mr Goodwill [35]

we'll start off by grouping some

\bf 3x^2-3x-5=13\implies (3x^2-3x)-5=13\implies 3(x^2-x)-5=13 \\\\\\ 3(x^2-x)=18\implies (x^2-x)=\cfrac{18}{3}\implies (x^2-x)=6\implies (x^2-x+~?^2)=6

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?

well, let's recall that a perfect square trinomial is

\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²

\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}

6 0
3 years ago
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