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Svet_ta [14]
3 years ago
5

Solve each of the following equations for x showing the algebra you used to do it. Round each answer to the nearest ten-thousand

s place.
11. 7^(x-3)=25

12. log (3x+1)

13. log(7x+1) = log(x-2) +1

14. 4e^(2x)-3=12

If you can't answer all of them that's totally fine. Thank you! ☻♥
Mathematics
1 answer:
lyudmila [28]3 years ago
4 0
11)46/7 thats the answer
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I think I know this much so far for A (but I could be wrong):
Eduardwww [97]
Part A. You have the correct first and second derivative.

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

Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.

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

Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out

To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h  ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0. 
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well. 
6 0
3 years ago
PLEASE HELP ME I WILL GIVE THE BRAINLIEST!!!!
Ipatiy [6.2K]

Answer:

9 cars/week

Step-by-step explanation:

(10+9+11+6)/4=36/4=9

8 0
3 years ago
Read 2 more answers
A recent poll of 124 randomly selected residents of a town with a population of 310 showed that 93 of them are opposed to a new
Stolb23 [73]
What is even the question?
7 0
2 years ago
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Prove that sin^4x + cos^4x= sinxcosx
yKpoI14uk [10]

Answer:

(the relation you wrote is not correct, there may be something missing, so I will simplify the initial expression)

Here we have the equation:

sin^4(x) + cos^4(x)

We can rewrite this as:

(sin^2(x))^2 + (cos^2(x))^2

Now we can add and subtract cos^2(x)*sin^2(x) to get:

(sin^2(x))^2 + (cos^2(x))^2 + 2*cos^2(x)*sin^2(x) - 2*cos^2(x)*sin^2(x)

We can complete squares to get:

(cos^2(x) + sin^2(x))^2 - 2*cos(x)^2*sin(x)^2

and we know that:

cos^2(x) + sin^2(x) = 1

then:

1 - 2*sin(x)^2*cos(x)^2

This is the closest expression to what you wrote.

We also know that:

sin(x)*cos(x) = (1/2)*sin(2*x)

If we replace that, we get:

1 - \frac{sin^2(2*x)}{2}

Then the simplification is:

cos^4(x) + sin^4(x)  = 1 - \frac{sin^2(2*x)}{2}

7 0
3 years ago
What two numbers can you multiply to get 84?
svet-max [94.6K]
84 ÷ 2 = 42

Check it to make sure.

2 × 42 = 84

42 and 2 can be multiplied to get 84.
7 0
3 years ago
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