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2 years ago

Which set makes the inequality m < -4 true

Mathematics

1 answer:

andriy [413]2 years ago

6 0

Answer:

To solve your inequality using the Inequality Calculator, type in your inequality like x+7>9. The inequality solver will then show you the steps to help you learn how to solve it on your own.

Step-by-step explanation:

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Please help with this

Degger [83]

All of them are wrong its 91 but if you want click or circle 90

3 0

4 years ago

The volume of the plastic cylinder is 64. A glass cylinder has the same height and radius is half of the plastic cylinder radius

sp2606 [1]

Answer:

I am not sure if I'm correct, but I got 16

Step-by-step explanation:

lets look at the formula for cylinder volume:

V=π*r^(2)*h

now, lets plug in 64

64=π*r^(2)*h

h=64/(π*r^(2))

Volume of glass cylinder is

V=((0.5r)^2)*(64/(π*r^(2)))*π

V=0.25π*r^(2)*(64/(π*r^(2)))

V=16π*r^(2)/(π*r^(2)

V=16

8 0

4 years ago

You have several purchases you

svetlana [45]

The answer: $24.05

How to:
Go through the graph and add all the values together
Then subtract $3.00 from the cost
Then add 4% sales tax

8 0

3 years ago

Find the volume of the specified cone. Use 3.14 for Pi, and round your answer to the nearest whole number. Radius = 5 in., Heigh

Irina-Kira [14]

Answer:

The answer is 236 cu. In.

Step-by-step explanation:

I just finished the practice quiz

7 0

3 years ago

Centerville is the headquarters of Greedy Cablevision Inc. The cable company is about to expand service to two nearby towns, Spr

NeX [460]

Answer:

20.2057 Units.

Step-by-step explanation:

First, we determine the length of the cable.

Distance between Centerville (8,0) and point (x,0) is given as:

$\sqrt{(8-x)}^2=8-x$

Distance between point (x,0) and Springfield(0,7) is:

$\sqrt{(7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}$

Distance between point (x,0) and Shelvyfield(0,-7) is:

$\sqrt{(-7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}$

Therefore the Length of the Cable L(x)

$L(x)=(8-x)+2\sqrt{(7)^2+(x)^{2}}$

To find the critical point, we set the derivative of L(x)=0

$L^{'}(x)=-1+\frac{2x}{\sqrt{\left( 49 - x^{2}\right) }}$

$\frac{-\sqrt{\left( 49 - x^{2}\right)}+2x}{\sqrt{\left( 49 - x^{2}\right)}}=0\\-\sqrt{\left( 49 - x^{2}\right)}+2x=0\\\sqrt{\left( 49 - x^{2}\right)}=2x\\(\sqrt{\left( 49 - x^{2}\right)})^2=(2x)^2\\49 - x^{2}=4x^2\\49=5x^2\\x^2=\frac{49}{5}\\x= 3.1305$

To verify that L(x) has a minimum at this critical number we compute the second derivative L″(x) and find its value at the critical number.

$L^{''}(x)=\frac{\left( 98\right) \,\sqrt{\left( 49 - x^{2}\right) }}{{\left( 7 - x\right) }^{2}\,{\left( 7+x\right) }^{2}}\\At \:x=3.1305, L^{''}=0.3993$

Since $L^{''}(x)$ is positive, the minimum point of L(x) exists.

Next, we find the minimum length by substituting z=3.1305 into L(x)

$L(3.1305)=(8-3.1305)+2\sqrt{(7)^2+(3.1305)^{2}}$

Minimum Length, L=20.2057

The minimum length of the cable is 20.2057 Units.

6 0

3 years ago