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OlgaM077 [116]
3 years ago
5

If logb3=0.23 and logb8=0.71, find logb24

Mathematics
1 answer:
prohojiy [21]3 years ago
6 0
Change logb24 into logb(8*3).
Next use the product rule to rewrite logb(8*3) as a sum.
logb8 + logb3
Now substitute the values given.
0.71 + 0.23 =
0.94
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Select the correct answer.
damaskus [11]

Answer:

None of the above.

Step-by-step explanation:

Let us check each of the answer choices one bye one.

Choice A: y+41=(31*2)+77x

solving for y we get:

y=77x+21,

which is not equal to our function h(x).

Choice B: y=317+77x+41.

This gives y=358+77x,

which is not equal to our function h(x).

Choice C: x+41=(31*2)+77x and y=31x+77x-41.

The first expression does not contain y, thus it is not equivalent to h(x). The second equation already gives the value of y , and we see that it is not equal to h(x).

Therefore we conclude that none of the choices given are correct.

4 0
4 years ago
Read 2 more answers
A cylinder has a volume of 864 in3 and a radius of 12 in. What is the height of the cylinder?
Firlakuza [10]
Cyl. vol.:  V = pi*r^2*h

Here,

V = 864 in^3 = pi*(12 in)^2*h

                     864 in^3
so that   h = --------------------- = 1.91 inches, or approx. 2 inches.
                      3.14*144 in^2
3 0
3 years ago
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Can you simplify a+2/5a^2+6*3/6
givi [52]
Your expression is ambiguous.

Do you mean a + (2/5)a^2 + 3?

or perhaps a + 2 / (5a^2) + 3?  Use parentheses for added clarity.

If you meant   a + (2/5)a^2 + 3, then

(2/5)a^2 + 1a + 3 (this can't be simplified further)
6 0
4 years ago
Susan is trying to decide if she should purchase renters insurance what would you say to advise her on this topic
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5 0
3 years ago
Aluminium coated dewars are cylindrical flasks used to safely store and transport liquid nitrogen, dry ice etc. A company manufa
Nitella [24]

This question is Incomplete because it lacks the diagram showing the containers as well as the required options. Find attached to this answer the appropriate diagram.

Complete Question

Aluminium coated dewars are cylindrical flasks used to safely store and transport liquid nitrogen, dry ice etc. A company manufactures such dewars of different sizes. Shown here are two dewars, both of the same height but of different diameters. How much aluminium will be required to cover Dewar A compared to that required for Dewar B? (Note: The base and the lid are NOT made up of aluminium.)

A) 1/3 times

B) 3 times

C) 9 times

D) The same amount of aluminum will be required.

Answer:

B) 3 times

Step-by-step explanation:

From the question, we are told that we are to find the amount of aluminum to cover Dewar A and B , we are also told that the base and the lid are not covered with aluminum.

From this information, we can determine that we are to find the curved or Lateral surface area of the cylinder because the base and lid are not included

The curved surface area of a cylinder is calculated as 2πrh

We told that both cylinders have the same height. Hence,

For Dewar A

We have Diameter of 30 cm, radius = Diameter ÷ 2 = 30cm ÷ 2 = 15cm.

Height = h

The curved surface area = 2πrh

= 2 × π × 15 ×h

= 30πhcm²

For Dewar B

We have Diameter of 10 cm, radius = Diameter ÷ 2 = 10cm ÷ 2 = 5cm.

Height = h

The curved surface area = 2πrh

= 2 × π × 5 ×h

= 10πhcm²

When we compare the curved surface Dewar A to the curved surface area of Dewar B

Dewar A : Dewar B

30πhcm² : 10πhcm²

3(10πhcm²) : 10πhcm²

From the above comparison, we can see that 3 times more aluminum is required to cover Dewar A compared to Dewar B.

6 0
3 years ago
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