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3 years ago

Please someone help me, i will give brainliest

Mathematics

1 answer:

raketka [301]3 years ago

4 0

Answer:

Both of them are the same measure: 45 degrees.

Step-by-step explanation:

You plug in 15 for the xin both of the equations they give.

3(15) +17=45

75-2(15)=45

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This is the right answer −4c+11

Step-by-step explanation:

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3 years ago

CAN SOMEONE HELP MEEEEEEEEEE

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$c = \sqrt{146} \\ x = 48°$

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Identify the slope and y-intercept of the graph of the equation.

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3 years ago

Read 2 more answers

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

CA Geometry A Illuminate Credit 4 FF.pdf<br> all 4 pls

lions [1.4K]

Answer:

1) HL=LJ=63

(ZJ)²=16²+63²

(ZJ)²=256+3969

ZJ=√4425

ZJ= 65

2)HJ=HL+LJ

= 63+63

HJ=126

3) XC=XA=8

4) BE=AB/2=20/2=10

<u>OAmalOHopeO</u>

6 0

3 years ago