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Black_prince [1.1K]
2 years ago
13

Find the quotient. (6x 2 + 23x + 20) ÷ (5 + 2x)

Mathematics
2 answers:
arsen [322]2 years ago
8 0

Answer:

The quotient is 3x+4

Step-by-step explanation:

Given  the dividend and divisor i.e we have to find the quotient if 6x^2+23x+20 is divided by 5+2x

Given dividend is

6x^2+23x+20

⇒ 6x^2+15x+8x+20

⇒ 3x(2x+5)+4(2x+5)

⇒ (3x+4)(2x+5)

Hence, the given dividend can written in two factors.

Now, \frac{6x^2+23x+20}{5+2x}=\frac{(3x+4)(2x+5)}{2x+5}=3x+4

Hence, the quotient is 3x+4

tekilochka [14]2 years ago
4 0
<span>(3x -4)(2x-5) as this it is divided </span>
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a container with a square bottom, rectangular sides and no top is to e constructed to have a volume of 5 m^3. material for the b
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Answer:

2m\times 2m\times \frac{5}{4}m

Step-by-step explanation:

We are given that a container with square bottom.

Let side of square=x

Height of container=h

Volume of container=5m^3

Cost of 1 square meter of material for the bottom=$10

Cost of 1 square meter of material for side=$8

We have to find the dimension of least expensive container.

Volume of container=x^2h

x^2h=5

h=\frac{5}{x^2}

Surface area of container=x^2+2(x+x)h=x^2+4xh

Cost=C(x)=10x^2+8(4xh)=10x^2+32x(\frac{5}{x^2})=10x^2+\frac{160}{x}

C(x)=10x^2+\frac{160}{x}

Differentiate w.r.t x

\frac{dC}{dx}=20x-\frac{160}{x^2}

\frac{dC}{dx}=0

20x-\frac{160}{x^2}=0

\frac{160}{x^2}=20x

x^3=\frac{160}{20}=8

x=\sqrt[3]{8}=2

Because side of container is always positive.

Again differentiate w.r.t x

\frac{d^2C}{dx^2}=20+\frac{320}{x^3}

Substitute x=2

\frac{d^2C}{dx^2}=20+\frac{320}{2^3}=60>0

Hence, the cost is minimum at x=2

Substitute the value of x

h=\frac{5}{2^2}=\frac{5}{4}m

Hence, the dimensions of the least expensive container are

2m\times 2m\times \frac{5}{4}m

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