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1 year ago

Solving and graphing absolute value equations

Mathematics

1 answer:

Monica [59]1 year ago

6 0

$\begin{gathered} -4\text{ + 7}x\text{ }\leq6x\text{ + 2} \\ 7x-6x\leq2+4 \\ x\text{ }\leq6 \end{gathered}$

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Find four consecutive odd integers such that 5 times the sum of the first and the third is 15 more than 5 times the fourth.

Dvinal [7]

Answer:

The required numbers are :

$\boxed{ \boxed{ 5,7,9 \: and \: 11}}$

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2 years ago

100 POINTS!!! Please solve this question step by step and explain each step for 100 points.

djverab [1.8K]

$2\dfrac 29 = -\dfrac 45 m \\\\\implies 2+ \dfrac 29 =- \dfrac 45 m \\\\\implies \dfrac{20}9 = -\dfrac 45 m \\\\\implies 5(20) = (-4m)9\\\\\implies 100 = -36m\\\\\implies m = -\dfrac{100}{36} \\\\\implies m = -\dfrac{25}9\\\\\implies m=-2\dfrac 79$

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2 years ago

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Use the given transformation to evaluate the integral

Misha Larkins [42]

Answer:

Step-by-step explanation:

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2 years ago

I need help relearning inequalities

Leto [7]

This is an example of how you solve an inequality.

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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