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3 years ago

7

At noon the temperature was 12 degrees and went up throughout the day. If the temperature was no more than 36 degrees. How many

degrees could the temperature have risen. Solve the inequality: 12 + x ≤ 36 *

1 answer:

kramer3 years ago

6 0

Answer:

B

Step-by-step explanation:

Hope this helps :D

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Find the volume of a regular hexahedron if one of the diagonals of its faces is 8 root 2 inches.

Naily [24]

The answer is 114sqrt{6} in³

A regular hexahedron is actually a cube.

Diagonal of a cube D is a hypotenuse of a right triangle which other two legs are face diagonal (f) and length of a side (a):
D² = f² + a²

Face diagonal is a hypotenuse of a right triangle which sides are a and a:
f² = a² + a² =2a²

D² = f² + a²
f² = 2a²

D² = 2a² + a² = 3a²
D = √3a² = √3 * √a² = √3 * a = a√3

Volume of a cube with side a is: V = a³
D = a√3
⇒ a = D/√3

V = a³ = (D/√3)³

We have:
D = 8√2 in

$V= (\frac{D}{ \sqrt{3} } )^{3} =( \frac{8 \sqrt{2} }{ \sqrt{3} } )^{3}=( \frac{8 \sqrt{2} * \sqrt{3} }{ \sqrt{3}* \sqrt{3}} )^{3} =( \frac{8 \sqrt{2*3} }{ \sqrt{3*3}} )^{3} =( \frac{8 \sqrt{6} }{ \sqrt{3^{2} }} )^{3} =( \frac{8 \sqrt{6} }{ 3} )^{3} = \\ \\ = \frac{ 8^{3} *( \sqrt{6} )^{3}}{3^{3}} = \frac{512* \sqrt{6 ^{2} }* \sqrt{6} }{27} = \frac{512*6* \sqrt{6} }{27} = 114sqrt{6}$

8 0

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I am thinking of two numbers. The sum of the numbers is 5. The first number is one less than the second number. Write a system o

Dominik [7]

Answer:

x + y = 5

y = x - 1

Hope this helped!

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What is the TOTAL area of a square with sides that are 67 cm and a rectangle with sides that are 47cm long and 27cm wide?

Mkey [24]

Answer:

94+54=148

Step-by-step explanation:

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3 years ago

Read 2 more answers

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Nuetrik [128]

The line is solid and under the line is shaded
so the answer is A

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Let μ denote the true average radioactivity level(picocuries per liter). The value 5 pCi/L is considered thedividing line betwee

VMariaS [17]

Answer:

H0: μ = 5 versus Ha: μ < 5.

Step-by-step explanation:

Given:

μ = true average radioactivity level(picocuries per liter)

5 pCi/L = dividing line between safe and unsafe water

The recommended test here is to test the null hypothesis, H0: μ = 5 against the alternative hypothesis Ha: μ < 5.

A type I error, is an error where the null hypothesis, H0 is rejected when it is true.

We know type I error can be controlled, so safer option which is to test H0: μ = 5 vs Ha: μ < 5 is recommended.

Here, a type I error involves declaring the water is safe when it is not safe. A test which ensures that this error is highly unlikely is desirable because this is a very serious error. We prefer that the most serious error be a type I error because it can be explicitly controlled.

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