What is the range of f(x)?

Answer:

cos(4) f'(sin 4)/f(sin 4) is y' at x=4

Step-by-step explanation:

We can't actually find the numerical value for this without more information.

We are given:

y=ln(f(sin x))

This is equivalent to

e^y=f(sin x)

Differentiate both sides:

y' e^y=cos(x) f'(sin x)

Divide both sides by e^y:

y' =cos(x) f'(sin x)/e^y

Rewrite in terms of x using that e^y=f(sin x):

y' =cos(x) f'(sin x)/f(sin x)

Now replace x with 4:

cos(4) f'(sin 4)/f(sin 4) is y' at x=4

Answer:

Imagine that scientists have just discovered a non-bird dinosaur skeleton. They want to know whether the dinosaur was closely related to birds. What features in ...

Step-by-step explanation:

If the sum of first four terms in a G.P. is 30 and that of the last four terms is 960, then the sum of the first 8 terms is 990

We also know that a = 2

sum of first 4 terms

= 2(r^4 - 1)(r-1) = 30

r^4 - 1 = 15r - 15

r^4 - 15r + 14 = 0

r = 1 works for the equation , but r = 1 would make all the terms of the

GP the same, so clearly not a solution

r^4 - 15r + 14 = 0 , a bit of trial and error ....

(r-1)(r - 2)(r^2 + 3r + 7) = 0

so r = 2 or r = not real

check: sum(4) = 2(2^4 - 1)/(2-1) = 30 , as needed

so r = 2 , BUT it does not work for sum(8)

sum(8) = 2(2^8 - 1)(2-1) = 510 , not 990 as necessary according to your second part of the question.

check your question.

Answer:

[tex]a^{2}[/tex]

Step-by-step explanation:

Answer: ?

Step-by-step explanation:

Answer:

an = a(n-1) -4

Step-by-step explanation:

From the table we see that

a1 = 15 , a2 = 11, a3 = 7 , a4= 3, a5 = -1...

Notice the numbers decrease by 4 so

a2 = a1 -4

a3 = a2 -4

a4 =a3 -4

a5= a4-4

generalized we will have

an = a n-1 -4

Answer:

were is the qn?

Step-by-step explanation:

PLACE VALUE. VALUE

6. Hundreds. 200

7. Ones. 1

8. Hundreds. 0

9. Hundreds. 800

10. Tens. 60

Answer:

999

Step-by-step explanation:

Answer:

999

Step-by-step explanation:

8% of 11,325 is 906

53 plus 40 is 93

906 plus 93 is 999

Answer:

-20%

Step-by-step explanation:

Current value of an item and divide it by the value from an earlier period. Subtract one and multiply the resulting number by 100 to give it a percentage representation.

ROC = (current value/previous value​ − 1) x 100

ROC = (1,000,000/1,250,000 - 1) x 100

ROC = (0.8 - 1) x 100

ROC = -0.2 x 100

ROC = -20

Determine whether a number is a solution to an equation.

- BRAINLIEST answerer

Answer:

substitute the value you solve the variable for back into the original equation and see if the left and right sides of the equation are equal

Step-by-step explanation:

let's say you have this equation:    

2x - 10 = 26

solve by first adding 10 to each side to get:

2x = 36

now divide each side by 2 to get:

x = 18

to check, substitute 18 in for 'x' in the original equation to see if the left side equals the right side of the equation:

2(18) - 10 = 26

36 - 10 = 26

26 = 26

we know our answer is correct

Answer:

i dont know sry....i ned points..

Answer: $180

Step-by-step explanation:

Answer:

180

Step-by-step explanation:

300-40% = 180 hshsjsbhshebs

Answer:

A

Step-by-step explanation:

The answer is A, because with absoulute value, the negative number will become positive. So positive 16 is obviously a bigger number than positive 4.

Have a great day friend! :D

Step-by-step explanation:

First, this is how I read it.

[tex]\log(x^{125})=-3[/tex]

Next, this is how logarithms work:

[tex]b^y=x\implies \log_b(x)=y[/tex]

In your equation, we have all 3 numbers:

With that information, you can convert the logarithm into "normal" form:

[tex]10^{-3}=x^{123}[/tex]

The negative power rule [tex]x^{-a}=\frac{1}{x^a}[/tex] is also useful here.

Now, just solve for x:

[tex]x^{125}=\frac{1}{10^3}\\x^{125}=\frac{1}{1000}\\x=\sqrt[125]{\frac{1}{1000}}\\x=\frac{1}{\sqrt[125]{1000} }[/tex]

That's about as simplified as it gets, unless you want this:

[tex]x=\frac{1}{10^{\frac{3}{125}}}[/tex]

Thats 10^(3/125), numbers got very small.

I'm still not confident I understood the problem correctly, so let me know if I need to fix anything.

T(2) = 3(2)³ - 2(2)² - 6(2) - 5

T(2) = 3(8) - 2(4) - 12 - 5

T(2) = 24 - 8 - 17

T(2) = 16 - 17

T(2) = -1 (B.)

Answer:

T = 4/5

Step-by-step explanation:

t/4 = 1/5

x4    x4

t = 4/5

Hope this works.

Answer:

4/5

Step-by-step explanation:

t/4 = 1/5

5t = 4

t = 4/5

yeah-ya.... right?

Answer: B

Solution: Because the line is a dotted line. so print (-3, -1) and (3, 3) , (0, 0) are not the solution, so , answer is (0, 5) choice B.

Answer:

C

Step-by-step explanation:

first way: (easier way)

Plot the points out, points in the blue region are solutions. points on the dotted line are not solutions since it is a dashed line.

(0,5 is in the blue region)

second way:

find the equation of the in-equality:

the line that divides solutions and non-solutions passes through 3,3 and -3,-1

y=mx+b

m=change in y/change in x = (-1-3)/(-3-3) = 2/3

y=2/3x+b

plug x=3, y=3 (from the point "3,3")

3=2/3*3+b

3=2+b

b=1

y=2/3x+1 is the solution of the line.

the in-equality is

y(?)2/3x+1

plug in -5,5, which is a solution (looking at the graph)

5>-7/3

so use >, and not <

y>2/3x+1 is the equation of the in-equality, now just plug the answer choices in to see that

x=0,y=5 is a solution

Answer: ?

Step-by-step explanation:

Answer:

wouldn't the answer be 9

Step-by-step explanation:

Answer:

22h^2 + 33 (added) .    h⋅h⋅22,2,22,2−3+10 (subtracted)

Step-by-step explanation:

I don't understand what you want so I will list lots of answers hopefully it helps. Please message under my answer for more solutions so I can help you but I hope this started to help you.

A is the correct answer

ok done. Thank to me :>

[tex]cosec18°= \frac{perpendicular}{base} \\ cosec18°= \frac{25}{x} \\ \\ \ \: since: cosec18°=0.9511 \\ \\ 0.9511 = \frac{25}{x} \\ x = \frac{25}{0.9} = 26[/tex]

Answer:

x=48 because 180-156=24. 108+24=132 180-132=48

Given that y varies inversely as (x-3) .When y is 5 x is 7 . We need to find out the value of y when x is 5 .

• According to the Question ,

[tex]\sf\longrightarrow y \propto \dfrac{1}{x-3}\\\\\sf\longrightarrow y =\dfrac{k}{x-3}[/tex]

And ,

[tex]\sf\longrightarrow 5 =\dfrac{k}{7}\\[/tex]

[tex]\sf\longrightarrow k = 7*5 \\ [/tex]

[tex]\bf\longrightarrow k = 35[/tex]

Now , when x is 5 ,

[tex]\sf\longrightarrow y = \dfrac{k}{x-3} \\ [/tex]

[tex]\sf\longrightarrow y =\dfrac{35}{5-3}\\ [/tex]

[tex]\sf\longrightarrow y=\dfrac{35}{2}\\ [/tex]

[tex]\sf\longrightarrow \boxed{\red{\sf y = 17.5 }} [/tex]