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Juliette [100K]
3 years ago
8

ted is twice his sister's age and three years younger than joe. ted's sister is half of a century plus ten. how old is ted how o

ld is ted's sister​
Mathematics
1 answer:
Alex17521 [72]3 years ago
4 0

Answer:

teds sister is 60 years old, ted is 120 some how, and joe is 123.

Step-by-step explanation:

half of a century is 50 because a century is 100 years and plus 10 years equals 60 and 60 times 60 equals 120 and 120 plus 3 years equals 123 so that's how I got the answers. At first this answer looks complicated you just have to read the question carefully.

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Find the product of √4(√8+3).<br> O 4√2+6√3<br> O 2√3+6<br> O 8√2+6√3<br> O 4√2+6
ella [17]

Answer:

Step-by-step explanation:

hello :

√4=2 and √8=2√2

√4(√8+3). =2(2√2+3)=4√2+6

4 0
2 years ago
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Write a decimal as a fraction simplest form for 7.75
Sonja [21]

Answer:

31/4

Step-by-step explanation:

Leave the 7 alone for now.

0.75 is equal to 75/100

so now simplfy 75/100=3/4

now add the 7 back in

its

7 and 3/4

That can be your final answer or you can make it improper by multiplying 4 * 7  + 3

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3 years ago
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Step-by-step explanation:

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3 years ago
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Find the absolute maximum and minimum values of the function below. f(x) = x3 − 9x2 + 3, − 3 2 ≤ x ≤ 12 Solution Since f is cont
Neko [114]

Answer:

There are an absolute minimum (x = 6) and an absolute maximum (x = 12).

Step-by-step explanation:

The correct statement is described below:

Find the absolute maximum and minimum values of the function below:

f(x) = x^{3}-9\cdot x^{2}+ 3, 2 \leq x \leq 12

Given that function is a polynomial, then we have the guarantee that function is continuous and differentiable and we can use First and Second Derivative Tests.

First, we obtain the first derivative of the function and equalize it to zero:

f'(x) = 3\cdot x^{2}-18\cdot x

3\cdot x^{2}-18\cdot x = 0

3\cdot x \cdot (x-6) = 0 (Eq. 1)

As we can see, only a solution is a valid critical value. That is: x = 6

Second, we determine the second derivative formula and evaluate it at the only critical point:

f''(x) = 6\cdot x -18 (Eq. 2)

x = 6

f''(6) = 6\cdot (6)-18

f''(6) =18 (Absolute minimum)

Third, we evaluate the function at each extreme of the given interval and the critical point as well:

x = 2

f(2) = 2^{3}-9\cdot (2)^{2}+3

f(2) = -25

x = 6

f(6) = 6^{3}-9\cdot (6)^{2}+3

f(6) = -105

x = 12

f(12) = 12^{3}-9\cdot (12)^{2}+3

f(12) = 435

There are an absolute minimum (x = 6) and an absolute maximum (x = 12).

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3 years ago
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Answer:

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