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Serga [27]
3 years ago
8

How to work this out in geometry form?

Mathematics
1 answer:
weeeeeb [17]3 years ago
8 0
B x H divided by 2 i think
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Two students, Carlos and Sophie, factored the trinomial -4x2 − 10x + 6. Sophie factored it as -2(2x − 1)(x + 3) and Carlos facto
damaskus [11]

Sophie has factored the term completely while Carlos has not factored the term completely.

Step-by-step explanation:

We need to factor the trinomial:

-4x^2-10x + 6

Factoring:

-4x^2-10x + 6\\Taking\,\,-2\,\,common:\\-2(2x^2+5x-3)\\Now\,\,factoring:\\-2(2x^2+6x-x-3)\\-2(2x(x+3)-1(x+3))\\-2(2x-1)(x+3)

So, the factored form of -4x^2-10x + 6 is -2(2x-1)(x+3)

So, Sophie has factored the term completely. The terms are factored when we cannot further simplify it. Since Carlos factored (x+3)(-4x+2) 2 can be taken common from the term (-4x+2) so, it is not factored completely.

Keywords: Factoring the terms

Learn more about Factoring the terms at:

  • brainly.com/question/1464739
  • brainly.com/question/1464739
  • brainly.com/question/2568692

#learnwithBrainly

7 0
3 years ago
Did I do this correct???
kondor19780726 [428]
The answer is actually D. Since 2.5 is a side of each tiny block, there is 3 sides that you are going to multiply together. There is 5,2 and 4 because we are multiplying the length,width and side. So the length is 2.5*5=12.5 which was correct. The width is 2.5*4=10 which you had but you crossed it out. Then lastly you are looking for the side of the shape. 2.5*2=5. Now that I have everything I just need to multiply everything together. 12.5*10*5=625 in³. I hope this helped you! 
6 0
3 years ago
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
Read 2 more answers
Can someone help mee :(((?
Shkiper50 [21]

Step-by-step explanation:

2√48= 13.86

3√147= 36.37

Perimeter: 13.86+13.86+36.37+36.37= 100.46

4 0
3 years ago
Could someone help me with number 6 ? The directions say what are the variables? which I know the variables are the height and t
kramer
As the time go the grass height go through increasing and decreasing
6 0
3 years ago
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