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2 years ago

Can anyone help me, I think we need to use desmos

Mathematics

1 answer:

Igoryamba2 years ago

5 0

Answer:

2 B l x Pattinson f s rıhtım dese rüyamın merhaba canım 53 nasılsın nasılsın bir yöntem

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Can someone answer plzzzz

telo118 [61]

Answer:

6,882,000 miles

Step-by-step explanation:

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2 years ago

The force in Newtons acting on a particle located x meters from the origin is given by the function F(x)=12x2+7. What is the wor

Shkiper50 [21]

$W = \displaystyle \int^{5}_{2} F(x) ~ dx\\\\\\= \displaystyle \int^{5}_{2} \left(12x^2 +7\right) ~dx\\\\\\= 12\displaystyle \int^{5}_{2} x^2 ~dx + \displaystyle \int^{5}_{2} 7~ dx\\\\\\=12 \left[\dfrac{x^3}3 \right]^{5}_{2} + 7\left[x\right]^{5}_{2}\\\\\\=4(5^3-2^3) + 7(5-2) = 468 + 21 = 489~ J$

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2 years ago

a rectangle is made by joining two squares adjacent to each other as shown below. if one squared has an area of 144 squared cent

liberstina [14]

Answer:

Hello! After reading your question I have deduced that the correct answer is 288² cm.

Step-by-step explanation:

The way I came to this conclusion was as follows:

Firstly:

If said rectangle is two squares put side by side (adjacent), then a valid assumption is that both squares are the same size.

This is because all four sides of a square have to be equal.

Thus if the two squares are joined together on one side, then all the other sides of both the squares will be the same length.

Thus both of the squares are going to be the same size, so they will have the same area.

Secondly:

If the area of one square is 144² cm then the area of the other square should also be 144² cm.

Thus if you combine the areas of both the squares, that make up the rectangle, you are left with the area of the rectangle being 288² cm.

I hope this helped!

3 0

2 years ago

Simplify . (1/c + 1/h)/(1/(c ^ 2) - 1/(r ^ 2))

Marrrta [24]

Answer:

$\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Step-by-step explanation:

$\frac{\frac{1}{c}+\frac{1}{h}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine $\frac{1}{c} + \frac{1}{h}$

$\frac{\frac{h+c}{ch}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine the bottom, too.

$=\frac{\frac{h+c}{ch}}{\frac{r^2-c^2}{c^2r^2}}$

Apply the fraction rule

$=\frac{\left(h+c\right)c^2r^2}{ch\left(r^2-c^2\right)}$

Cancel

$=\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Therefore, $\frac{\left(\frac{1}{c}+\frac{1}{h}\right)}{\left(\frac{1}{\left(c^2\right)}-\frac{1}{\left(r^2\right)}\right)}:\quad \frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

5 0

3 years ago

Can someone please help me with these questions?

tensa zangetsu [6.8K]

Answer:

c and a i think

Step-by-step explanation:

7 0

2 years ago