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Shtirlitz [24]
3 years ago
10

Solve the given problem. Lizzie is 2 years older than three times her sister Clarissa’s age. If Clarissa is 4 years old, how old

is Lizzie?
****PLZ HELP****
Mathematics
2 answers:
umka2103 [35]3 years ago
8 0
The correct answer is 14!
valina [46]3 years ago
7 0

Answer:

14. 4 times 3 is 12, 12 plus 2 is 14.

Step-by-step explanation:


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What is the solution for -2 [3.25 + 1/2 divided by 1/4 x (-6)]
Gelneren [198K]

Answer:

-4.5

Step-by-step explanation:

I added 3.25 and .5=3.75. Then i did 0.25*-6=1.5. -1.5+3.75=2.25. 2.25*2=-4.5

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Describe what the inequality x ≥ 3 looks like when graphed
Pachacha [2.7K]

<, > - open circle

≤, ≥ - closed circle

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>, ≥ - draw to the right

--------------------------------------------

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borishaifa [10]

Answer:

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Step-by-step explanation:

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2 years ago
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find a curve that passes through the point (1,-2 ) and has an arc length on the interval 2 6 given by 1 144 x^-6
taurus [48]

Answer:

f(x) = \frac{6}{x^2} -8 or f(x) = -\frac{6}{x^2} + 4

Step-by-step explanation:

Given

(x,y) = (1,-2) --- Point

\int\limits^6_2 {(1 + 144x^{-6})} \, dx

The arc length of a function on interval [a,b]:  \int\limits^b_a {(1 + f'(x^2))} \, dx

By comparison:

f'(x)^2 = 144x^{-6}

f'(x)^2 = \frac{144}{x^6}

Take square root of both sides

f'(x) =\± \sqrt{\frac{144}{x^6}}

f'(x) = \±\frac{12}{x^3}

Split:

f'(x) = \frac{12}{x^3} or f'(x) = -\frac{12}{x^3}

To solve fo f(x), we make use of:

f(x) = \int {f'(x) } \, dx

For: f'(x) = \frac{12}{x^3}

f(x) = \int {\frac{12}{x^3} } \, dx

Integrate:

f(x) = \frac{12}{2x^2} + c

f(x) = \frac{6}{x^2} + c

We understand that it passes through (x,y) = (1,-2).

So, we have:

-2 = \frac{6}{1^2} + c

-2 = \frac{6}{1} + c

-2 = 6 + c

Make c the subject

c = -2-6

c = -8

f(x) = \frac{6}{x^2} + c becomes

f(x) = \frac{6}{x^2} -8

For: f'(x) = -\frac{12}{x^3}

f(x) = \int {-\frac{12}{x^3} } \, dx

Integrate:

f(x) = -\frac{12}{2x^2} + c

f(x) = -\frac{6}{x^2} + c

We understand that it passes through (x,y) = (1,-2).

So, we have:

-2 = -\frac{6}{1^2} + c

-2 = -\frac{6}{1} + c

-2 = -6 + c

Make c the subject

c = -2+6

c = 4

f(x) = -\frac{6}{x^2} + c becomes

f(x) = -\frac{6}{x^2} + 4

3 0
2 years ago
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