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2 years ago

Please help! Read the question below!

Mathematics

2 answers:

Sunny_sXe [5.5K]2 years ago

7 0

Answer:1 quarter, 2 dimes and 1 nickel, 5 nickels, 3 nickels and 1 dime

Step-by-step explanation:

Ksivusya [100]2 years ago

5 0

Sorry but I’m answering this so I can post my question :)

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Solve the inequality x/5-3<0

Kisachek [45]

Answer:

x < 15

Step-by-step explanation:

$\frac{x}{5}$ - 3 < 0 ( add 3 to both sides )

$\frac{x}{5}$ < 3 ( multiply both sides by 5 to clear the fraction )

x < 15

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2 years ago

Seventh grade

kupik [55]

-5+1+-2+9v+-5v

-3+1+4v

-2+4v

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2 years ago

Solve this quadratic equation using the quadratic formula. 5 - 10x - 3x2 = 0

horsena [70]

Answer:

x = (-5 ± 2√10) / 3

Step-by-step explanation:

5 − 10x − 3x² = 0

Write in standard form:

-3x² − 10x + 5 = 0

Solve with quadratic formula:

x = [ -b ± √(b² − 4ac) ] / 2a

x = [ -(-10) ± √((-10)² − 4(-3)(5)) ] / 2(-3)

x = [ 10 ± √(100 + 60) ] / -6

x = (10 ± 4√10) / -6

x = (-5 ± 2√10) / 3

4 0

3 years ago

Read 2 more answers

A ball of radius 15 has a round hole of radius 5 drilled through its center. Find the volume of the resulting solid. Hint: The u

OverLord2011 [107]

Answer:

The volume of the ball with the drilled hole is:

$\displaystyle\frac{8000\pi\sqrt{2}}{3}$

Step-by-step explanation:

See attached a sketch of the region that is revolved about the y-axis to produce the upper half of the ball. Notice the function y is the equation of a circle centered at the origin with radius 15:

$x^2+y^2=15^2\to y=\sqrt{225-x^2}$

Then we set the integral for the volume by using shell method:

$\displaystyle\int_5^{15}2\pi x\sqrt{225-x^2}dx$

That can be solved by substitution:

$u=225-x^2\to du=-2xdx$

The limits of integration also change:

For x=5: $u=225-5^2=200$

For x=15: $u=225-15^2=0$

So the integral becomes:

$\displaystyle -\int_{200}^{0}\pi \sqrt{u}du$

If we flip the limits we also get rid of the minus in front, and writing the root as an exponent we get:

$\displaystyle \int_{0}^{200}\pi u^{1/2}du$

Then applying the basic rule we get:

$\displaystyle\frac{2\pi}{3}u^{3/2}\Bigg|_0^{200}=\frac{2\pi(200\sqrt{200})}{3}=\frac{400\pi(10)\sqrt{2}}{3}=\frac{4000\pi\sqrt{2}}{3}$

Since that is just half of the solid, we multiply by 2 to get the complete volume:

$\displaystyle\frac{2\cdot4000\pi\sqrt{2}}{3}$

$=\displaystyle\frac{8000\pi\sqrt{2}}{3}$

5 0

3 years ago

HELP MEEE!! What's the domain and range?

r-ruslan [8.4K]

Answer:

domain: [- ∞ , ∞]

range: (8 , -∞]

Step-by-step explanation:

- ∞ < x < ∞

y < 8

7 0

2 years ago