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dimulka [17.4K]
3 years ago
13

The Carolina Castaways and Raleigh Rockets scored a total of 96 points. The Rockets scored one third as many points as the Casta

ways.
Mathematics
2 answers:
zysi [14]3 years ago
8 0
The castaways scored 32 pts

Hope it helps
11Alexandr11 [23.1K]3 years ago
4 0
I believe it's 32
96/1(1/3)
96/3
32
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Find the volume of the composite solid. Round your answer to the nearest tenth.
EastWind [94]

Answer:

646.04 cm³

Step-by-step explanation:

<u>Cube volume formula:</u>

V=l*w*h

V=8*8*8

V=512

<u>Hemisphere volume formula:</u>

V=(4/3πr³)1/2

V=(4/3π4³)1/2

V=128/3

<u>Add both together</u>

128/3 + 512 ≈ 646.04cm³

4 0
3 years ago
Find the y-intercept of the line that has a slope of -3 and passes through (9,0)
Ann [662]

9514 1404 393

Answer:

  27

Step-by-step explanation:

For point (x, y) and slope m, the y-intercept can be found from the equation ...

  b = y -mx

  b = 0 -(-3)(9) = 27

The y-intercept is +27. The corresponding ordered pair is (0, 27).

7 0
2 years ago
How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?
Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+...  \ \ \  \ \ --- (1)

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)          

f'(x) = \dfrac{1}{1+x}

f'(0)   = \dfrac{1}{1+0}=1

f ' ' (x)    = \dfrac{1}{(1+x)^2}

f ' ' (x)   = \dfrac{1}{(1+0)^2}=-1

f '  ' '(x)   = \dfrac{2}{(1+x)^3}

f '  ' '(x)    = \dfrac{2}{(1+0)^3} = 2

f ' '  ' '(x)    = \dfrac{6}{(1+x)^4}

f ' '  ' '(x)   = \dfrac{6}{(1+0)^4}=-6

f ' ' ' ' ' (x)    = \dfrac{24}{(1+x)^5} = 24

f ' ' ' ' ' (x)    = \dfrac{24}{(1+0)^5} = 24

Now, the next process is to substitute the above values back into equation (1)

f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \  '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...

In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...

In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...

To estimate the value of In(1.4), let's replace x with 0.4

In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...

In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...

Therefore, from the above calculations, we will realize that the value of \dfrac{0.4^5}{5}= 0.002048 as well as \dfrac{0.4^6}{6}= 0.00068267 which are less than 0.001

Hence, the estimate of In(1.4) to the term is \dfrac{0.4^5}{5} is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0
3 years ago
I need help with problem 19
Vsevolod [243]

Answer:

124

Step-by-step explanation:

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2 years ago
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What is 494 x 20.62 rounded to the hundredths place?
DENIUS [597]
10186.28
And that’s how you do it
8 0
3 years ago
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