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spayn [35]
2 years ago
9

It's simplifying radicals. the problem is I thought I knew how to do this and turns out I don't. If anyone would be wiling to le

nd a hand. That would be much appreciated.

Mathematics
1 answer:
dmitriy555 [2]2 years ago
4 0

Here is a image of his contact options.

here is a video too.

https://youtu.be/zfmNx_sC82A

I hope this helps.

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What is the answer to -7d-22=4d?
Juliette [100K]
First add 7d to both sides:
-22 = 11d
Now divide by 11 both sides
-2 = d

You can work on either side of the equation, I used the right side for the variables because it was easier to add to positive numbers than dealing with negative numbers on the left side:
 -7d -4d = 22
-11d =22
d = -2 

either way you get the same answer :)
4 0
3 years ago
4 over 3 devided by 2?
Veronika [31]

Answer:

2/3 or 2 over 3

Step-by-step explanation:

4 over three means 4 pieces out of 3. 4 pieces divided by 2 is 2 pieces.

8 0
2 years ago
Read 2 more answers
Suppose an airline policy states that all baggage must be box shaped with a sum of​ length, width, and height not exceeding 114
NISA [10]

Answer:

Step-by-step explanation:

Represent the length of one side of the base be s and the height by h.  Then the volume of the box is V = s^2*h; this is to be maximized.

The constraints are as follows:  2s + h = 114 in.  Solving for h, we get 114 - 2s = h.

Substituting 114 - 2s for h in the volume formula, we obtain:

V = s^2*(114 - 2s), or V = 114s^2 - 2s^3, or V = 2*(s^2)(57 - s)

This is to be maximized.  To accomplish this, find the first derivative of this formula for V, set the result equal to 0 and solve for s:

dV

----- = 2[(s^2)(-1) + (57 - s)(2s)] = 0 = 2s^2(-1) + 114s - 2s^2

ds

Simplifying this, we get dV/ds = -4s^2 + 114s = 0.  Then either s = 28.5 or s = 0.

Then the area of the base is 28.5^2 in^2 and the height is 114 - 2(28.5) = 57 in

and the volume is V = s^2(h) = 46,298.25 in^3

7 0
3 years ago
Expand the following using the Binomial Theorem and Pascal’s triangle. (x + 2)6 (x − 4)4 (2x + 3)5 (2x − 3y)4 In the expansion o
ivolga24 [154]
\bf (2x+3)^5\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(2x)^5(+3)^0\\
2&+5&(2x)^4(+3)^1\\
3&+10&(2x)^3(+3)^2\\
4&+10&(2x)^2(+3)^3\\
5&+5&(2x)^1(+3)^4\\
6&+1&(2x)^0(+3)^5
\end{array}

as you can see, the terms exponents, for the first term, starts at highest, 5 in this case, then every element it goes down by 1, till it gets to 0

for the second term, starts at 0, and every element it goes up by 1, till it gets to the highest

now, to get the coefficient, they way I get it, is "the product of the current coefficient and the exponent of the first term, divided by the exponent of the second term plus 1"

notice the first coefficient is always 1

so...how did we get 10 for the 3rd element?  well, 5*4/2
how did we get 10 for the fourth element?  well, 10*2/4


\bf (2x-3y)^4\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(2x)^4(-3y)^0\\
2&+4&(2x)^3(-3y)^1\\
3&+6&(2x)^2(-3y)^2\\
4&+4&(2x)^1(-3y)^3\\
5&+1&(2x)^0(-3y)^4
\end{array}


\bf (3a+4b)^8\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(3a)^8(+4b)^0\\
2&+8&(3a)^7(+4b)^1\\
3&+28&(3a)^6(+4b)^2\\
4&+56&(3a)^5(+4b)^3\\
5&+70&(3a)^4(+4b)^4\\
6&+56&(3a)^3(+4b)^5\\
7&+28&(3a)^2(+4b)^6\\
8&+8&(3a)^1(+4b)^7\\
9&+1&(3a)^0(+4b)^8
\end{array}

and from there, you can simplify the elements of the expansion by combining the coefficients

like for example, the 7th element of (3a+4b)⁸ will then be 1032192a²b⁶
7 0
3 years ago
Read 2 more answers
HELP NEEDED PLEASE!!!! This math question confuses me. I thought the line didn't pass through any other point but I cant figure
Vanyuwa [196]

Answer:

B

Step-by-step explanation:

So wiggly, remember that slope is change in y/change in x, or in simpler terms, y/x. Thinking of this, y/x must simplyfy to -3. Or, you can think of it as -3/1.

So know that we have rewrote this as a slope of -3/1, lets use it. Where do we start? Well, we must start at point A, which is at 2 x and 7 y.

Now, going to the  RIGHT(+1x) and going DOWN(-3y) we will go from 2x and 7y to 3x and 4y.

Are we on a point yet? Nope, lets do this again.

Now going to RIGHT again(+1x) and going DOWN again(-3y) we will go from 13x and 4y to 4x and 1y.

Are we on a point yet? Still no, lets try this one more time.

Now going to RIGHT again(+1x) and going DOWN again(-3y) we will go from 4x and 1y to 5x and -2y.

Are we on a point? YUP. We cna see that point B is at 5x and -2y.

Hope you get it right ;)

4 0
3 years ago
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