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frosja888 [35]
3 years ago
11

Solve for x: 1 over 2 plus 4 over 2 x equals quantity x plus 4 over 10

Mathematics
2 answers:
qaws [65]3 years ago
7 0
\frac{1}{2} + \frac{4}{2x} = \frac{x + 4}{10}
\frac{x}{2x} + \frac{4}{2x} = \frac{x + 4}{10}
\frac{x + 4}{2x} = \frac{x + 4}{10}
10(x + 4) = 2x(x + 4)
10(x) + 10(4) = 2x(x) + 2x(4)
10x + 40 = 2x^{2} + 8x
40 = 2x^{2} - 2x
20 = x^{2} - x
0 = x^{2} - x - 20
0 = (x - 5)(x + 4)
0 = x - 5\ or\ 0 = x + 4
5 = x\ or\ -4 = x
Fynjy0 [20]3 years ago
7 0

Answer:

there are two solutions when adding your fraction- your two answers should be x=-4 and x=5, which is answer choice C for this problem! hope this helped you!

Step-by-step explanation:


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Please help I want summer school done. Which graph corresponds to the given function?
Alexus [3.1K]

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Step-by-step explanation:

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2 years ago
Can y’all plz help me ?
crimeas [40]

Answer:

x=-5

Step-by-step explanation

The first box has eight x's and six 1's which means that it would be written as 8x+6

The second box has four x's and fourteen -1's which means that it would be written as 4x-14

You set them equal to one another

8x+6=4x-14

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4x+6=-14

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4 0
3 years ago
A rectangular box with a square base is to be constructed from material that costs $9 per ft2 for the bottom, $5 per ft2 for the
kramer

Answer:

18.73ft^3

Step-by-step explanation:

Let

Side of square base=x

Height of rectangular box=y

Area of square base=Area of top=(side)^2=x^2

Area of one side face=l\times b=xy

Cost of bottom=$9 per square ft

Cost of top=$5 square ft

Cost of sides=$4 per square ft

Total cost=$204

Volume of rectangular box=V=lbh=x^2y

Total cost=9(x^2)+5x^2+4(4xy)=14x^2+16xy

204=14x^2+16xy

204-14x^2=16xy

y=\frac{204-14x^2}{16x}=\frac{102-7x^2}{8x}

Substitute the values of y

V(x)=x^2\times \frac{102-7x^2}{8x}=\frac{1}{8}(102x-7x^3)

Differentiate w.r.t x

V'(x)=\frac{1}{8}(102-21x^2)=0

V'(x)=0

\frac{1}{8}(102-21x^2)=0

102-21x^2=0

102=21x^2

x^2=\frac{102}{21}=4.85

x=\sqrt{4.85}=2.2

It takes positive because side length cannot be negative.

Again differentiate w.r. t x

V''(x)=\frac{1}{8}(-42x)

Substitute the value

V''(2.2)=-\frac{42}{8}(2.2)=-11.55

Hence, the volume of box is maximum at x=2.2 ft

Substitute the value  of x

y=\frac{102-7(2.2)^2}{8(2.2)}=3.87 ft

Greatest volume of box=x^2y=(2.2)^2\times 3.87=18.73 ft^3

5 0
3 years ago
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