3 years ago

Being timed need help

Mathematics

2 answers:

Contact [7]3 years ago

8 0

B)acute scalene let me know if it’s wrong sorry if I am

andreyandreev [35.5K]3 years ago

6 0

Answer:

it is the last option right isosceles

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Rearrange to make x the subject of 4(x-3)/a=y

IrinaVladis [17]

Answer: x = a*y + 3

Step-by-step explanation:

To make x the subject of the equation, first, we open the bracket

4x - 12/a = y

Then cross multiply:

4x - 12 = a * y ( a*y means the product of the two variables)

Add 12 to both sides of the equation

4x = a*y + 12

Divide both sides by 4 to get the value of x

x = a * y + 12/4

x = a*y + 3

I hope this helps.

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3 years ago

Ticketd to the school play cost $11 for adults and $6 for children. What expression can be used to find the toyal cost for x adu

Makovka662 [10]

Answer:221

Step-by-step explanation:

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3 years ago

A particle moves along the x-axis so that at any time t, t ≥ 0, its acceleration is a(t) = -4sin(2t). If the velocity of the par

sasho [114]

The velocity of the particle is given by

$v(t)=v(0)+\displaystyle\int_0^ta(u)\,\mathrm du$

Since $a(t)=-4\sin2t$ and $v(0)=7$ , we get

$\displaystyle\int_0^ta(u)\,\mathrm du=\int_0^t-4\sin2u\,\mathrm du=2\cos2u\bigg|_0^t=2\cos2t-2$

$\implies v(t)=2\cos2t+5$

Similarly, the position function is obtained via

$x(t)=x(0)+\displaystyle\int_0^tv(u)\,\mathrm du$

We know $v(t)$ and we're told that $x(0)=0$ , so

$\displaystyle\int_0^tv(u)\,\mathrm du=\int_0^t(2\cos2u+5)\,\mathrm du=\sin2u+5u\bigg|_0^t=\sin2t+5t$

$\implies x(t)=\sin2t+5t$

making the answer A.

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3 years ago

Help!!!!!<br> factor 3^2+15x-42

nydimaria [60]

Answer:

3(5x−11)

Step-by-step explanation:

32+15x−42

15x−33

=3(5x−11)

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2 years ago

Whats the area of the shape?

Karolina [17]

I could be wrong but i think it’s 32

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2 years ago

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