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sineoko [7]
3 years ago
12

Write the expression that represents twice the sum of x and 5

Mathematics
1 answer:
posledela3 years ago
7 0
Answer:

2(x+5)

Step by Step explanation:
Since sum means the answer of an addition problem, that means it would be x+5. and since it is twice that, it would be 2*x+5, which we would then write as 2(x+5) so it could be easier to read and understand.
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WILL GIVE BRAINLIEST WHAT IS THE FORMULA FOR SURFACE AREA
eimsori [14]

Answer:

Hi! The formula for surface area is:

area = 2lw + 2lh + 2hw

(l = length, w = width, h = height.)

<em>Hope this helps! :)</em>

8 0
3 years ago
Read 2 more answers
A vendor has 16 balloons for sale 18 are yellow 3 are red and 5 are green.A balloon is selected at random and sold .if the ballo
Elena L [17]

Well since there are more Yellow balloons there is a less chance of it getting picked twice because there other colors.

So it will probably be 7 over 15

6 0
3 years ago
Find the domain of the function y = 3 tan(23x)
solmaris [256]

Answer:

\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

In other words, the x in f(x) = 3\, \tan(23\, x) could be any real number as long as x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) for all integer k (including negative integers.)

Step-by-step explanation:

The tangent function y = \tan(x) has a real value for real inputs x as long as the input x \ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

Hence, the domain of the original tangent function is \mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

On the other hand, in the function f(x) = 3\, \tan(23\, x), the input to the tangent function is replaced with (23\, x).

The transformed tangent function \tan(23\, x) would have a real value as long as its input (23\, x) ensures that 23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2} for all integer k.

In other words, \tan(23\, x) would have a real value as long as x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right).

Accordingly, the domain of f(x) = 3\, \tan(23\, x) would be \mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)  \; \right| k \in \mathbb{Z}  \right\rbrace.

4 0
2 years ago
A student committee is to consist of 2 freshmen, 5 sophomores, 4 juniors, and 3 seniors. If 6 freshmen, 13 sophomores, 8 juniors
mezya [45]
<h3>Answer:  491,891,400</h3>

Delete the commas if necessary.

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

Explanation:

There are 6 freshmen total and we want to pick 2 of them, where order doesn't matter. The reason it doesn't matter is because each seat on the committee is the same. No member outranks any other. If the positions were labeled "president", "vice president", "secretary", etc, then the order would matter.

Plug n = 6 and r = 2 into the nCr combination formula below

n C r = \frac{n!}{r!(n-r)!}\\\\6 C 2 = \frac{6!}{2!*(6-2)!}\\\\6 C 2 = \frac{6!}{2!*4!}\\\\6 C 2 = \frac{6*5*4!}{2!*4!}\\\\ 6 C 2 = \frac{6*5}{2!}\\\\ 6 C 2 = \frac{6*5}{2*1}\\\\ 6 C 2 = \frac{30}{2}\\\\ 6 C 2 = 15\\\\

This tells us there are 15 ways to pick the 2 freshmen from a pool of 6 total.

Repeat those steps for the other grade levels.

n = 13 sophomores, r = 5 selections leads to nCr = 13C5 = 1287. This is the number of ways to pick the sophomores.

You would follow the same type of steps shown above to get 1287. Let me know if you need to see these steps.

Similarly, 8C4 = 70 is the number of ways to pick the juniors.

Lastly, 14C3 = 364 is the number of ways to pick the seniors.

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

To recap, we have...

  • 15 ways to pick the freshmen
  • 1287 ways to pick the sophomores
  • 70 ways to pick the juniors
  • 364 ways to pick the seniors

Multiply out those values to get to the final answer.

15*1287*70*364 = 491,891,400

This massive number is a little under 492 million.

7 0
2 years ago
In the coordinate plane, the directed line segment from K to N has endpoints at K(–6, –2) and N(8, 3). Point L partitions the di
Dafna1 [17]

its c .i just took it


7 0
3 years ago
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