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3 years ago

The number line shows values of x that make the inequality x > 1 true.

Mathematics

2 answers:

Lana71 [14]3 years ago

5 0

Where is the number line?

luda_lava [24]3 years ago

4 0

Number line please? Like where is it

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Please help me now I’m a beginner and I so need help please explain how did you get you answer so I understand thanks

Irina-Kira [14]

Answer:y=2.5(x)

Step-by-step explanation:y=2.5(x)

y=2.5(2)= y=5

7 0

3 years ago

WILL GIVE BRAINLIEST

scoray [572]

Inverse Variation is shown through the formula y=k/x
plug in the given values and solve for k: 3=k/5 so k=15

Answer: y=15/x

4 0

3 years ago

All possible roots of x^3+2x^2-16x-32

nekit [7.7K]

Answer:

See below.

Step-by-step explanation:

By the rational roots theorem all possible roots are:

+/-1. +/- 2, +/-4 .+/-8, +/-16 and +/-32.

4 0

3 years ago

Read 2 more answers

What is the approximate area of a 70 decree sector of a circle with a radius of 8 inches

puteri [66]

The area of the sector will also be 70/360
Sector area = 70/360 • pir^2
= 7/36 • pi • 8^2
A= 112/9 • pi
A= 39.1 inch^2

3 0

3 years ago

A rectangular box is to have a square base and a volume of 12 ft3. If the material for the base costs $0.17/ft2, the material fo

katen-ka-za [31]

Answer:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

Step-by-step explanation:

Let the dimensions of the box be x, y and z

The rectangular box has a square base.

Therefore, Volume of the box $V=x^2z$

Volume of the box $=12 ft^3\\$

$Therefore, x^2z=12\\z=\frac{12}{x^2}$

The material for the base costs $\$0.17/ft^2$ , the material for the sides costs $\$0.10/ft^2$ , and the material for the top costs $\$0.13/ft^2$ .

Area of the base $=x^2$

Cost of the Base $=\$0.17x^2$

Area of the sides $=4xz$

Cost of the sides= $=\$0.10(4xz)$

Area of the Top $=x^2$

Cost of the Base $=\$0.13x^2$

Total Cost, $C(x,z) =0.17x^2+0.13x^2+0.10(4xz)$

Substituting $z=\frac{12}{x^2}$

$C(x) =0.17x^2+0.13x^2+0.10(4x)(\frac{12}{x^2})\\C(x)=0.3x^2+\frac{4.8}{x} \\C(x)=\dfrac{0.3x^3+4.8}{x}$

To minimize C(x), we solve for the derivative and obtain its critical point

$C'(x)=\dfrac{0.6x^3-4.8}{x^2}\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8\div 0.6\\x^3=8\\x=\sqrt[3]{8}=2$

Recall: $z=\frac{12}{x^2}=\frac{12}{2^2}=3\\$

Therefore, the dimensions that minimizes the cost of the box are:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

7 0

3 years ago