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stepladder [879]
3 years ago
10

Find b 33^2+b^2=66^2

Mathematics
1 answer:
viktelen [127]3 years ago
7 0
<span>The answer is:
b=<span><span><span>33<span>√3</span></span><span> or </span></span>b</span></span>=<span>−<span>33<span>√<span>3 </span></span></span></span>
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35/40 reducing fractions
laila [671]

Answer: 7/8

What you have to do here is to find the highest number which can be divided by both. Or you have to divide both the numerator and denominator until it can not be divided by any number anymore. You can do this by finding the highest common factor (<u>HCF)</u> of both the number, 35 and 40.

Highest Common Factor: HCF of two or more numbers is the greatest factor that divides the numbers. For example, 2 is the HCF of 4 and 6.

So, the highest number which divides 35 and 40, is 5. Now 35 divided by 5 is 7 and 40 divided by 5 is 8. The final answer is 7/8 and there is no more numbers which can divide both of them by a specific number.


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   hope it helped
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6 0
2 years ago
X/8-1/2=6 <br>what does that make?​
slega [8]

Answer:

<h2>              x = 10</h2>

Step-by-step explanation:

\frac x8-\frac12\ =\ 6\\\\{}\ \ \cdot8\qquad \cdot8\\\\x-4\ =\ 6\\\\{}\ +4\ \,\quad+4\\\\x\ =\ 10

5 0
3 years ago
How do you find the absolute extrema of this function?<br> f(x) = x+3x^(2/3); Interval is [-10,1]
Step2247 [10]
\bf f(x)=x+3x^{\frac{2}{3}}\implies \cfrac{dy}{dx}=1+3\left(\frac{2}{3}x^{-\frac{1}{3}}  \right)\implies \cfrac{dy}{dx}=1+\cfrac{2}{\sqrt[3]{x}}&#10;\\\\\\&#10;\cfrac{dy}{dx}=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\cfrac{\sqrt[3]{x}+2}{\sqrt[3]{x}}\implies 0=\sqrt[3]{x}+2\implies -2=\sqrt[3]{x}&#10;\\\\\\&#10;(-2)^3=x\implies \boxed{-8=x}\\\\&#10;-------------------------------\\\\&#10;0=\sqrt[3]{x}\implies \boxed{0=x}

now, f(0) = 0, and f(-8) is an imaginary value or no real value.

now, f(-10)   will also give us an imaginary value

and f(1) = 4

so, doing a first-derivative test on 0, is imaginary to the left and positive on the right, and before and after 1, is positive as well, so f(x) is going up on those intervals.

however, f(0) is 0 and f(1) is higher up, so the absolute maximum will have to be f(1), and we can use f(0) as a minimum, and since it's the only one, the absolute minimum.

the other two, the endpoint of -10 and the critical point of -8, do not yield any values for f(x).
8 0
3 years ago
How would you write 9^4 as an equivalent expression using a base of 3
valina [46]

Answer:

why u should go f u cc k urself

Step-by-step explanation:

Cuz u got a big p e n iis

8 0
4 years ago
Read 2 more answers
Over which interval is the function f(x)=x^3-27/x^2-9 differentiable?
aksik [14]

Answer:

Step-by-step explanation:

F(x)=x³-27/x²-9

F(x)=(-9)³-27/(-9)²-9

F(x)=-729-27/-81-9

F(x)=-756/-90

F(x)=42/5

8 0
3 years ago
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