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4 years ago

A candle is 6 inches tall and burns at a rate of 1/2 inch per hour. How much would be left of the candle after 3 hours?

Mathematics

1 answer:

aleksandr82 [10.1K]4 years ago

8 0

The candle would be 4.5 inches tall after three hours.

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Let’s now add x to both sides of the equation and consider the new equation x squared+5+x=11

irina1246 [14]

Answer:x=3

Step-by-step explanation:x squared is one x.you then add the X's to get 2x.then you subtract the five on the other side to get 2x=6.after you get that you divide both sides by 2 to get x=3

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3 years ago

Explanation. Please Help!!!

Svetach [21]

Answer:

The domain is y > 3

Step-by-step explanation:

y = 2⁻ˣ + 3

As x approaches -∞, y approaches ∞.

As x approaches ∞, y approaches 3.

So the range is y > 3, and there is an asymptote at y = 3.

Compared to the parent function y = 2ˣ, it is reflected over the y-axis and shifted up 3 units.

The false statement is "the domain is y > 3". Domain describes the possible values of x. Range describes the possible values of y.

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3 years ago

The expression (secx + tanx)2 is the same as _____.

trapecia [35]

<u>Answer:</u>

The expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

<u>Solution:</u>

From question, given that $\bold{(\sec x+\tan x)^{2}}$

By using the trigonometric identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$ the above equation becomes,

$(\sec x+\tan x)^{2} = \sec ^{2} x+2 \sec x \tan x+\tan ^{2} x$

We know that $\sec x=\frac{1}{\cos x} ; \tan x=\frac{\sin x}{\cos x}$

$(\sec x+\tan x)^{2}=\frac{1}{\cos ^{2} x}+2 \frac{1}{\cos x} \frac{\sin x}{\cos x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

$=\frac{1}{\cos ^{2} x}+\frac{2 \sin x}{\cos ^{2} x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

On simplication we get

$=\frac{1+2 \sin x+\sin ^{2} x}{\cos ^{2} x}$

By using the trigonometric identity $\cos ^{2} x=1-\sin ^{2} x$ ,the above equation becomes

$=\frac{1+2 \sin x+\sin ^{2} x}{1-\sin ^{2} x}$

By using the trigonometric identity $(a+b)^{2}=a^{2}+2ab+b^{2}$

we get $1+2 \sin x+\sin ^{2} x=(1+\sin x)^{2}$

$=\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}$

$=\frac{(1+\sin x)(1+\sin x)}{1-\sin ^{2} x}$

By using the trigonometric identity $a^{2}-b^{2}=(a+b)(a-b)$ we get $1-\sin ^{2} x=(1+\sin x)(1-\sin x)$

$=\frac{(1+\sin x)(1+\sin x)}{(1+\sin x)(1-\sin x)}$

$= \frac{1+\sin x}{1-\sin x}$

Hence the expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

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3 years ago

The distribution of waiting time for an elevator in an office building is shown above. What is the probability you’ll have to sp

Oxana [17]

Answer:

I don't think this question can be answered without more info

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3 years ago

I can’t seem to figure out how I would graph this? Someone help please

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3 years ago

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