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2 years ago

Find the domain and range of the following graph Write your answer as an interval.

Mathematics

1 answer:

HACTEHA [7]2 years ago

8 0

Answer:

Domain: [1, infinity)

Range: All real numbers or (- infinity, + infinity)

Step-by-step explanation:

Hope this helps

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What does it mean when a researcher states that an observed effect or difference is significant with a confidence level of 99%?

icang [17]

Answer:

it affected more than half

Step-by-step explanation:

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3 years ago

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One of the legs of a right triangle is twice as long as the other, and the perimeter of the triangle is 35. Find the

Aloiza [94]

Answer:

6.68, 13.37, 14.95

Step-by-step explanation:

One of the legs is twice as long as the other.

b = 2a

The perimeter is 35.

35 = a + b + c

The triangle is a right triangle.

c² = a² + b²

Three equations, three variables. Start by plugging the first equation into the second and solving for c.

35 = a + 2a + c

c = 35 − 3a

Now plug this and the first equation into the Pythagorean theorem:

(35 − 3a)² = a² + (2a)²

1225 − 210a + 9a² = a² + 4a²

1225 − 210a + 4a² = 0

Solve with quadratic formula:

a = [ -(-210) ± √((-210)² − 4(4)(1225)) ] / 2(4)

a = (210 ± √24500) / 8

a ≈ 6.68 or 45.82

Since the perimeter is 35, a = 6.68. Therefore, the other sides are:

b ≈ 13.37

c ≈ 14.95

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3 years ago

The hour hand of a clock moves -15 degrees every hour how many degrees does it move in 1 1/4 hours

iren [92.7K]

Answer:

-18 3/4

Step-by-step explanation:

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3 years ago

Which three statements describe how to find the product of 5×10^3

Vedmedyk [2.9K]

5x10x1 electrical and electrical equipment for electrical

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3 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

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3 years ago