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2 years ago

Can someone help me??

Mathematics

1 answer:

pshichka [43]2 years ago

7 0

Answer:

(8,8)

Step-by-step explanation:

The first equation says y and x are the same, so any x in the second equation can be replaced with a y, so

y = 2x - 8 and y = x, substituting x for y gives

x = 2x -8, subtract x from each side, add 8 to each side, get

8 = x

substitute that value back into the first equation

y = x, x= 8, so y=8

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Solve the equation by completing the square. Round to the nearest hundredth if necessary. x2 – x – 7 = 0

Tanzania [10]

2x-x-7=0

3x-7=0

add 7 to 7 and 0, so then we cross out 7-7, and then

3x=7, bc u change the subtraction to addition, sr if thst doesn't make sense but 3x=7 is the answer

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A local television station runs 2.5 minutes of commercials during every 30 minutes of programming. At this rate, how many minute

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12.5 minutes of commercials

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2 years ago

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Angle ABC is congruent to which angle?

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CBA=MNP

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2 years ago

Ax +by=c solve for y

hichkok12 [17]

Answer:

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Step-by-step explanation:

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Hint: | Isolate terms with y to the left hand side.

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2 years ago

A closed, rectangular-faced box with a square base is to be constructed using only 36 m2 of material. What should the height h a

kkurt [141]

Answer:

$b=h=\sqrt{6}$ m

Step-by-step explanation:

Let

Bas length of box=b

Height of box=h

Material used in constructing of box=36 square m

We have to find the height h and base length b of the box to maximize the volume of box.

Surface area of box= $2b^2+4bh$

$2b^2+4bh=36$

$b^2+2bh=18$

$2bh=18-b^2$

$h=\frac{18-b^2}{2b}$

Volume of box, V= $b^2h$

Substitute the values

$V=b^2\times \frac{18-b^2}{2b}$

$V=\frac{1}{2}(18b-b^3)$

Differentiate w. r.t b

$\frac{dV}{db}=\frac{1}{2}(18-3b^2)$

$\frac{dV}{db}=0$

$\frac{1}{2}(18-3b^2)=0$

$\implies 18-3b^2=0$

$\implies 3b^2=18$

$b^2=6$

$b=\pm \sqrt{6}$

$b=\sqrt{6}$

The negative value of b is not possible because length cannot be negative.

Again differentiate w.r.t b

$\frac{d^2V}{db^2}=-3b$

At $b=\sqrt{6}$

$\frac{d^2V}{db^2}=-3\sqrt{6}$

Hence, the volume of box is maximum at $b=\sqrt{6}$ .

$h=\frac{18-(\sqrt{6})^2}{2\sqrt{6}}$

$h=\frac{18-6}{2\sqrt{6}}$

$h=\frac{12}{2\sqrt{6}}$

$h=\sqrt{6}$

$b=h=\sqrt{6}$ m

7 0

3 years ago