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2 years ago

Ma= P2-P1/t solve for p1

Mathematics

1 answer:

natulia [17]2 years ago

3 0

Answer:

P1 = P2 - ma*t

Step-by-step explanation:

ma= P2-P1/t

we multiply by t both sides of the equation

ma*t = (P2 - P1)*t/t

ma*t = P2 - P1

we sum by P1 both sides of the equation:

P1 +ma*t = P2 - P1 +P1

We ave:

P1 + ma*t = P2

we subtract by ma*t both sides of the equation:

P1 + ma*t -ma*t = P2 - ma*t

finally we have:

P1 = P2 - ma*t

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The formula for the volume V of a cylinder is V = πr2h, where r is the radius of the base and h is the height of the cylinder. S

lana [24]

Answer:

Part A) $h=V/(\pi r^{2})$

Part B) $h=4\ cm$

Step-by-step explanation:

Part A)

we have the formula of the volume of a cylinder

$V=\pi r^{2}h$

Solve for h

That means ----> isolate the variable h

Divide both side by πr²

$V/(\pi r^{2})=\pi r^{2}h/(\pi r^{2})$

Simplify

$V/(\pi r^{2})=h$

Rewrite

$h=V/(\pi r^{2})$

Part B) Find the height of a cylinder with a volume of 36π cm3 and a base with a radius of 3 cm

we have

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substitute in the formula and solve for h

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2 years ago

What are the solutions of the equation (2x + 3)2 + 8(2x + 3) + 11 = 0? Use u substitution and the quadratic formula to solve.

zvonat [6]

Answer:

$x=-2.38$

$x=-4.62$

Step-by-step explanation:

The question is $(2x+3)^2+8(2x+3)+11=0$

We let $u=2x+3$ , so the equation becomes:

$u^2+8u+11=0$

Where $a=1, b=8, c=11$

Putting it in the quadratic formula, we have:

Quadratic formula: $\frac{-b+-\sqrt{b^2-4ac} }{2a}$

Substituting we have: $\frac{-8+-\sqrt{(8)^2-4(1)(11)} }{2(1)}\\=\frac{-8+-\sqrt{20} }{2}\\=\frac{-8+-2\sqrt{5} }{2}\\=-4+\sqrt{5}, -4-\sqrt{5} }$

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$u=2x+3\\(-4+\sqrt{5})=2x+3\\x=\frac{-7+\sqrt{5}}{2}=-2.38$

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The solutions of the equation is $x=-2.38$ (rounded to 2 decimal places), and $x=-4.62$ (rounded to 2 decimal places)

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2 years ago

Convert 778.5 million km to scientific notation.

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2 years ago

Ma= P2-P1/t solve for p1​

Ma= P2-P1/t solve for p1