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zvonat [6]
3 years ago
6

Use the Pythagorean Theorem to find the hypotenuse.

Mathematics
1 answer:
valkas [14]3 years ago
4 0
Is the hypotenuse not already shown? wouldn’t it just be 10…?

if your trying to find the value of the missing leg then that would 8.
10^2-6^2= b
100-36=64
square root of 64 is 8.


not sure if this was a trick question or just bad wording cause the longest leg- the hypotenuse is already shown in that image-10. otherwise if it was bad wording and means the missing leg, then that would be 8.

so based off the question it’s 10. it’s just weird how they said to use the pythagorean theorem when you don’t have to…
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PLEASE HELP, GOOD ANSWERS GET BRAINLIEST. +40 POINTS WRONG ANSWERS GET REPORTED
MA_775_DIABLO [31]
1. Ans:(A) 123

Given function: f(x) = 8x^2 + 11x
The derivative would be:
\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2 + 11x)
=> \frac{d}{dx} f(x) = \frac{d}{dx}(8x^2) + \frac{d}{dx}(11x)
=> \frac{d}{dx} f(x) = 2*8(x^{2-1}) + 11
=> \frac{d}{dx} f(x) = 16x + 11

Now at x = 7:
\frac{d}{dx} f(7) = 16(7) + 11

=> \frac{d}{dx} f(7) = 123

2. Ans:(B) 3

Given function: f(x) =3x + 8
The derivative would be:
\frac{d}{dx} f(x) = \frac{d}{dx}(3x + 8)
=> \frac{d}{dx} f(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(8)
=> \frac{d}{dx} f(x) = 3*1 + 0
=> \frac{d}{dx} f(x) = 3

Now at x = 4:
\frac{d}{dx} f(4) = 3 (as constant)

=>Ans:  \frac{d}{dx} f(4) = 3

3. Ans:(D) -5

Given function: f(x) = \frac{5}{x}
The derivative would be:
\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{5}{x})
or 
\frac{d}{dx} f(x) = \frac{d}{dx}(5x^{-1})
=> \frac{d}{dx} f(x) = 5*(-1)*(x^{-1-1})
=> \frac{d}{dx} f(x) = -5x^{-2}

Now at x = -1:
\frac{d}{dx} f(-1) = -5(-1)^{-2}

=> \frac{d}{dx} f(-1) = -5 *\frac{1}{(-1)^{2}}
=> Ans: \frac{d}{dx} f(-1) = -5

4. Ans:(C) 7 divided by 9

Given function: f(x) = \frac{-7}{x}
The derivative would be:
\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{-7}{x})
or 
\frac{d}{dx} f(x) = \frac{d}{dx}(-7x^{-1})
=> \frac{d}{dx} f(x) = -7*(-1)*(x^{-1-1})
=> \frac{d}{dx} f(x) = 7x^{-2}

Now at x = -3:
\frac{d}{dx} f(-3) = 7(-3)^{-2}

=> \frac{d}{dx} f(-3) = 7 *\frac{1}{(-3)^{2}}
=> Ans: \frac{d}{dx} f(-3) = \frac{7}{9}

5. Ans:(C) -8

Given function: 
f(x) = x^2 - 8

Now if we apply limit:
\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 - 8)

=> \lim_{x \to 0} f(x) = (0)^2 - 8
=> Ans: \lim_{x \to 0} f(x) = - 8

6. Ans:(C) 9

Given function: 
f(x) = x^2 + 3x - 1

Now if we apply limit:
\lim_{x \to 2} f(x) = \lim_{x \to 2} (x^2 + 3x - 1)

=> \lim_{x \to 2} f(x) = (2)^2 + 3(2) - 1
=> Ans: \lim_{x \to 2} f(x) = 4 + 6 - 1 = 9

7. Ans:(D) doesn't exist.

Given function: f(x) = -6 + \frac{x}{x^4}
In this case, even if we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

Check:
f(x) = -6 + \frac{x}{x^4} \\ f(x) = -6 + \frac{1}{x^3} \\ f(x) = \frac{-6x^3 + 1}{x^3} \\ Rationalize: \\ f(x) = \frac{-6x^3 + 1}{x^3} * \frac{x^{-3}}{x^{-3}} \\ f(x) = \frac{-6x^{3-3} + x^{-3}}{x^0} \\ f(x) = -6 + \frac{1}{x^3} \\ Same

If you apply the limit, answer would be infinity.

8. Ans:(A) Doesn't Exist.

Given function: f(x) = 9 + \frac{x}{x^3}
Same as Question 7
If we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

9, 10.
Please attach the graphs. I shall amend the answer. :)

11. Ans:(A) Doesn't exist.

First We need to find out: \lim_{x \to 9} f(x) where,
f(x) = \left \{ {{x+9, ~~~~~x \textless 9} \atop {9- x,~~~~~x \geq 9}} \right.

If both sides are equal on applying limit then limit does exist.

Let check:
If x \textless 9: answer would be 9+9 = 18
If x \geq 9: answer would be 9-9 = 0

Since both are not equal, as 18 \neq 0, hence limit doesn't exist.


12. Ans:(B) Limit doesn't exist.

Find out: \lim_{x \to 1} f(x) where,

f(x) = \left \{ {{1-x, ~~~~~x \textless 1} \atop {x+7,~~~~~x \textgreater 1} } \right. \\ and \\ f(x) = 8, ~~~~~ x=1

If all of above three are equal upon applying limit, then limit exists.

When x < 1 -> 1-1 = 0
When x = 1 -> 8
When x > 1 -> 7 + 1 = 8

ALL of the THREE must be equal. As they are not equal. 0 \neq 8; hence, limit doesn't exist.

13. Ans:(D) -∞; x = 9

f(x) = 1/(x-9).

Table:

x                      f(x)=1/(x-9)       

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

8.9                       -10

8.99                     -100

8.999                   -1000

8.9999                 -10000

9.0                        -∞


Below the graph is attached! As you can see in the graph that at x=9, the curve approaches but NEVER exactly touches the x=9 line. Also the curve is in downward direction when you approach from the left. Hence, -∞,  x =9 (correct)

 14. Ans: -6

s(t) = -2 - 6t

Inst. velocity = \frac{ds(t)}{dt}

Therefore,

\frac{ds(t)}{dt} = \frac{ds(t)}{dt}(-2-6t) \\ \frac{ds(t)}{dt} = 0 - 6 = -6

At t=2,

Inst. velocity = -6


15. Ans: +∞,  x =7 

f(x) = 1/(x-7)^2.

Table:

x              f(x)= 1/(x-7)^2     

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

6.9             +100

6.99           +10000

6.999         +1000000

6.9999       +100000000

7.0              +∞

Below the graph is attached! As you can see in the graph that at x=7, the curve approaches but NEVER exactly touches the x=7 line. The curve is in upward direction if approached from left or right. Hence, +∞,  x =7 (correct)

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