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forsale [732]
3 years ago
7

Which interval is the solution set to 0.35x – 4.8 < 5.2 – 0.9x

Mathematics
2 answers:
Daniel [21]3 years ago
8 0
<span>0.35x – 4.8 < 5.2 – 0.9x
0.35x + 0.9x < 5.2 + 4.8
1.25x < 10
x < 10/1.25
x < 8

Solutions are all real numbers (-∞, 8)
</span>
EastWind [94]3 years ago
8 0

Answer:

B) (–∞, 8)

Step-by-step explanation:

So the interval for the solution of a certain set is the numbers that can be used as value for X in the set and the set still be true, in order to solve this, you just need to solve the set by making it an equation:

0.35x – 4.8 < 5.2 – 0.9x

0.35x – 4.8 =5.2 – 0.9x

0.35x+0.9x=5.2+4.8

1.25x=10

x=8

So the maximum value of X will be 8, and when X is a negative value the set is true, so the interval would be all real numbers form -infinite to 8.

(-∞,8)

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The ratio of volume of a cone and a hemisphere of same base radius and height is?
aliya0001 [1]

Answer:

1/2

Step-by-step explanation:

Since it is given that the cone and the hemisphere have the same height, and since the height of a hemisphere would be equal to its radius, the cones height must also be equal to its base radius.

With this information we can use the respective volume formulas.

Hemisphere:  \frac{2}{3}πr^3

Cone:  \frac{1}{3}πhr^2

Since h (height) = r we can say the cone volume equals:

\frac{1}{3}πr^3

Now to find the ratio we divide the cone volume equation by the hemisphere volume equation

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7 0
3 years ago
Prove the trigonometric identity
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Answer:

Proved See below

Step-by-step explanation:

Man this one is a world of its own :D Just a quick question are you a fellow Add Math student in O levels i remember this question from back in the day :D Anyhow Lets get started

For this question we need to know the following identities:

1+tan^{2}x=sec^2x\\\\1+cot^2x=cosec^2x\\\\sin^2x+cos^2x=1

Lets solve the bottom most part first:

1-\frac{1}{1-sec^2x} \\\\

Take LCM

1-\frac{1}{1-sec^2x} \\\\\frac{1-sec^2x-1}{1-sec^2x} \\\\\frac{-sec^2x}{1-sec^2x} \\\\\frac{-(1+tan^2x)}{-tan^2x}

now break the LCM

\frac{-1}{-tan^2x}+\frac{-tan^2x}{-tan^2x}\\\\\frac{1}{tan^2x}+1\\\\cot^2x+1

because 1/tan = cot x

and furthermore,

cot^2x+1\\cosec^2x

now we solve the above part and replace the bottom most part that we solved with cosec^2x

\frac{1}{1-\frac{1}{cosec^2x} } \\\\\frac{1}{1-sin^2x} \\\\\frac{1}{cos^2x}\\\\sec^2x

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4 0
3 years ago
Setting:
nata0808 [166]
A. Mr. Kent interviewed the 54 students as they are going to leave the school, it is not considered to be a random sample. It is because a random sample is when a set is taken from a population. Mr. Kent interviewed the 54 who are going to leave, meaning, he didn't take a set out of that 54, he took all of them. So it is not a random sample.

b. The question that Mr. Kent asked is considered to be a leading question, so it does not seem biased.

c. If there are 54 respondents.
51 = yes, the rest is no.
= 54 - 51 = 3
= 3 is now divided to 54 = 3/54
= giving an answer of 0.0555
= 0.0555 x 100
= 5.6%
= The percent of responses that says 'no' is 5.6%

6 0
4 years ago
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Its height would be 7 meters.

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