if f(x)=3^x+10x and g(x)=5x-3, find (f-g)(x)

Answer:

0.1, 0.2, 1/2

Answer:

0.1<0.2<0.5(1/2)

hope it helps

have a nice day

Answer:

[tex]{ \tt{ \frac{( \frac{x + y}{x - y}) }{( \frac{y + x}{y - x}) } }} \\ \\ { \tt{ = \frac{x + y}{x - y} \times \frac{y - x}{y + x} }} \\ \\ { \tt{ = \frac{-(x- y)}{x - y} }}[/tex]

Answer: = -1

Answer:

Option B

Step-by-step explanation:

By applying the converse theorem of corresponding angles,

"If corresponding angles formed between two parallel lines and the transversal line are equal then both the lines will be parallel"

Angle between line B and Y = 90°

Angle between line B and Z = 90°

Therefore, corresponding angles are equal.

By applying converse theorem, line Y and line Z will be parallel.

Option B will be the answer.

Answer:

a) 41.666...%

b) 10/120 or 1/12

Step-by-step explanation:

50 students like volleyball and tennis out of 120 students surveyed. 50/120 as a percent is 41.6666666...%

10 students like volleyball but not tennis out of 120 students surveyed. 10/120 can also simplify to 1/12

:) ur welcome

Answer:

x = 82

Step-by-step explanation:

1/3(12/7 + 9/7) + 3⁴ =

1/3(21/7) + 3⁴ =

1/3(3) + 3⁴ =

1 + 81 = 82

Answer:

4.61 years

Step-by-step explanation:

hope it helped!

Answer:

The correct option is (b).

Step-by-step explanation:

The formula for the side of a cube of surface area SA is as follows :

[tex]s=\sqrt{\dfrac{SA}{6}}[/tex]

When SA = 180 m²

[tex]s=\sqrt{\dfrac{180}{6}}\\\\s=\sqrt{30}[/tex]

When SA = 120 m²

[tex]s=\sqrt{\dfrac{120}{6}}\\\\s=\sqrt{20}\\\\=2\sqrt5[/tex]

Difference,

[tex]=30-2\sqrt5[/tex]

So, the correct option is (b).

Answer:

-4/5

Step-by-step explanation:

If you use the number line, after adding 3/5 you can see that it still doesn't make it positive but brings it up to -4/5 (Hope this helps)

Answer:

You subtract them:

3/5-7/5 (the plus disappears when faced with a minus)

-4/5

Answer:

8.

a) f'x means you find the derivative.

2 * d/dx x^2 -b * d/dx x + d/dx c

use power rule x^2 = 2x^1

2*2x = 4x.  the derivative of the differentiation variable, x is 1 and the derivative of a constant, c is 0

4x-b+0

4x-b is our derivative

(I am still figuring out b and c, I will edit this answer and put the solution for b and c.)

Step-by-step explanation:

Answer:

[tex] \displaystyle - \frac{1}{2} [/tex]

Step-by-step explanation:

we would like to compute the following limit:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{1}{ \ln(x + \sqrt{ {x}^{2} + 1} ) } - \frac{1}{ \ln(x + 1) } \right) [/tex]

if we substitute 0 directly we would end up with:

[tex] \displaystyle\frac{1}{0} - \frac{1}{0} [/tex]

which is an indeterminate form! therefore we need an alternate way to compute the limit to do so simplify the expression and that yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right) [/tex]

now notice that after simplifying we ended up with a rational expression in that case to compute the limit we can consider using L'hopital rule which states that

[tex] \rm \displaystyle \lim _{x \to c} \left( \frac{f(x)}{g(x)} \right) = \lim _{x \to c} \left( \frac{f'(x)}{g'(x)} \right) [/tex]

thus apply L'hopital rule which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \ln(x + 1) - \ln(x + \sqrt{ {x}^{2} + 1 } }{ \dfrac{d}{dx} \ln(x + \sqrt{ {x}^{2} + 1} ) \ln(x + 1) } \right) [/tex]

use difference and Product derivation rule to differentiate the numerator and the denominator respectively which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \frac{1}{x + 1} - \frac{1}{ \sqrt{x + 1} } }{ \frac{ \ln(x + 1)}{ \sqrt{ {x}^{2} + 1 } } + \frac{ \ln(x + \sqrt{x ^{2} + 1 } }{x + 1} } \right) [/tex]

simplify which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \sqrt{ {x}^{2} + 1 } - x - 1 }{ (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right) [/tex]

unfortunately! it's still an indeterminate form if we substitute 0 for x therefore apply L'hopital rule once again which yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \dfrac{d}{dx} \sqrt{ {x}^{2} + 1 } - x - 1 }{ \dfrac{d}{dx} (x + 1)\ln(x + 1 ) + \sqrt{ {x}^{2} + 1} \ln( x + \sqrt{ {x }^{2} + 1} ) } \right) [/tex]

use difference and sum derivation rule to differentiate the numerator and the denominator respectively and that is yields:

[tex] \displaystyle \lim _{x \to 0} \left( \frac{ \frac{x}{ \sqrt{ {x}^{2} + 1 } } - 1}{ \ln(x + 1) + 2 + \frac{x \ln(x + \sqrt{ {x}^{2} + 1 } ) }{ \sqrt{ {x}^{2} + 1 } } } \right) [/tex]

thank god! now it's not an indeterminate form if we substitute 0 for x thus do so which yields:

[tex] \displaystyle \frac{ \frac{0}{ \sqrt{ {0}^{2} + 1 } } - 1}{ \ln(0 + 1) + 2 + \frac{0 \ln(0 + \sqrt{ {0}^{2} + 1 } ) }{ \sqrt{ {0}^{2} + 1 } } } [/tex]

simplify which yields:

[tex] \displaystyle - \frac{1}{2} [/tex]

finally, we are done!

9514 1404 393

Answer:

  -1/2

Step-by-step explanation:

Evaluating the expression directly at x=0 gives ...

  [tex]\dfrac{1}{\ln(\sqrt{1})}-\dfrac{1}{\ln(1)}=\dfrac{1}{0}-\dfrac{1}{0}\qquad\text{an indeterminate form}[/tex]

Using the linear approximations of the log and root functions, we can put this in a form that can be evaluated at x=0.

The approximations of interest are ...

  [tex]\ln(x+1)\approx x\quad\text{for x near 0}\\\\\sqrt{x+1}\approx \dfrac{x}{2}+1\quad\text{for x near 0}[/tex]

__

Then as x nears zero, the limit we seek is reasonably approximated by the limit ...

  [tex]\displaystyle\lim_{x\to0}\left(\dfrac{1}{x+\dfrac{x^2}{2}}-\dfrac{1}{x}\right)=\lim_{x\to0}\left(\dfrac{x-(x+\dfrac{x^2}{2})}{x(x+\dfrac{x^2}{2})}\right)\\\\=\lim_{x\to0}\dfrac{-\dfrac{x^2}{2}}{x^2(1+\dfrac{x}{2})}=\lim_{x\to0}\dfrac{-1}{2+x}=\boxed{-\dfrac{1}{2}}[/tex]

_____

I find a graphing calculator can often give a good clue as to the limit of a function.

Answer:

C

Step-by-step explanation:

Given

24x² - 22x + 5

Consider the factors of the product of the x² term and the constant term which sum to give the coefficient of the x- term.

product = 24 × 5 = 120 and sum = - 22

The factors are - 12 and - 10

Use these factors to split the x- term

24x² - 12x - 10x + 5 ( factor the first/second and third/fourth terms )

= 12x(2x - 1) + 5(2x - 1)  ← factor out (2x - 1) from each term

= (12x - 5)(2x - 1) ← in factored form → C

Answer:

a

Step-by-step explanation:

Answer:

a. 0.1536

b. 0.9728

Step-by-step explanation:

The probability that a component fails, P(Y) = 0.2

The number of components in the system = 4

The number of components required for the subsystem to operate = 2

a. By binomial theorem, we have;

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) is given as follows;

P(Y = 2) = [tex]\dbinom{4}{2}[/tex] × 0.2² × 0.8² = 0.1536

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) = 0.1536

b. The probability that the system last longer than 1,000 hours, P(O) = The probability that no component fails + The probability that only one component fails + The probability that two component fails leaving two working

Therefore, we have;

P(O) = P(Y = 0) + P(Y = 1) + P(Y = 2)

P(Y = 0) = [tex]\dbinom{4}{0}[/tex] × 0.2⁰ × 0.8⁴ = 0.4096

P(Y = 1) = [tex]\dbinom{4}{1}[/tex] × 0.2¹ × 0.8³ = 0.4096

P(Y = 2) = [tex]\dbinom{4}{2}[/tex] × 0.2² × 0.8² = 0.1536

∴ P(O) = 0.4096 + 0.4096 + 0.1536 = 0.9728

The probability that the subsystem operates longer than 1,000 hours = 0.9728

Answer:

width?

8-W-52

2(w-8) +2w=52

(W-8)+W=52

2(8-W) +2w=52

Answer:

It is not a solution

Step-by-step explanation:

Plug the point into the equations and check to see if they are true

y=3x - 3

3 = 3(6) -3

3 = 18-3

3 = 15

False

We do not need to check the other equation since this is false

Question options:

A. He should report them directly on form 1040

B. He should report them on form 8949 and then on schedule D

C. He should report them on schedule D

D. He is not required to report them until he sells the underlying securities

Answer:

B. He should report them on form 8949 and then on schedule D

Explanation:

John has shares which have capital gains from a mutual fund and a brokerage account. In order to report his taxes, he would need to use the Schedule D(form 1040) for his mutual fund capital gains and the form 8949 for his brokerage capital gains. The brokerage capital gains is then transferred to schedule D.

Answer:

[tex]\frac{1+3(n-1)}{3+4(n-1)}[/tex]

Step-by-step explanation:

First find the pattern.

1/3 gets 5/21 added to it but doing that to 4/7 does not get 7/11, so it's nto normal adding.

if you multiply 1/3 gets 12/7, but if you multiply that by 4/7 you don't get 7/11 so it's not normal multiplcation.

I would next try only adding to the numerator and denominator separately.  

so 1/3 gets 3 added to the 1 and 4 added to the 3.  Doing that again gets us (4+3)/(7+4) = 7/11 and doing it again gets us 10/15 = 2/3.  So that is the right answer.

So we know what is happening.  if you start with 1/3 and increasingthe numerator by 3 and denominator by 4 then we know it's going to look like (1+3(n-1))/(3+4(n-1)) because the first term  is when n=1 and we want that to cancel out.  You can also simplify it and get (3n-2)/(4n-1)

Let me know if it doesn't make sense.  

Answer:

87 minutes

Step-by-step explanation:

Let the total number of minutes = m

Our equation is given as:

$20.21 = $14.99 + 0.06m

20.21 = 14.99 + 0.06m

Collect like terms

0.06m= 20.21 - 14.99

0.06m = 5.22

m = 5.22/0.06

m = 87

Therefore, the total number rod minutes used is 87 minutes

Answer:

500000 cm2

Step-by-step explanation:

Answer:

6th term = 25

Step-by-step explanation:

3n + 7

3 x (6) + 7

18 + 7 = 25

If this helps you, please mark brainliest!

Have a nice day!

Answer:

[tex]25[/tex]

Step-by-step explanation:

[tex] Tn_{n} = 3n + 7 \\Tn _{6} = 3 n + 7 \\ = 3 \times 6 + 7 \\ = 18 + 7 \\ = 25[/tex]

Hope this helps you

Have a nice day!

Answer:  [tex]2\sqrt{2}+\sqrt{3}\\\\[/tex]

a = 2 and b = 1

=======================================================

Explanation:

Set the expression equal to the given form we want. Then square both sides so we get rid of the outer-most square root

[tex]\sqrt{11+4\sqrt{6}} = a\sqrt{2}+b\sqrt{3}\\\\\left(\sqrt{11+4\sqrt{6}}\right)^2 = \left(a\sqrt{2}+b\sqrt{3}\right)^2\\\\11+4\sqrt{6} = \left(a\sqrt{2}\right)^2+2*a\sqrt{2}*b\sqrt{3}+\left(b\sqrt{3}\right)^2\\\\11+4\sqrt{6} = 2a^2+2ab\sqrt{2*3}+3b^2\\\\11+4\sqrt{6} = 2a^2+3b^2+2ab\sqrt{6}\\\\[/tex]

In the third line, I used the rule that (x+y)^2 = x^2+2xy+y^2

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

At this point, we equate the non-radical and radical terms to get this system of equations

[tex]\begin{cases}11 = 2a^2+3b^2\\ 4\sqrt{6} = 2ab\sqrt{6}\end{cases}[/tex]

The second equation turns into 4 = 2ab when we divide both sides by sqrt(6)

Then 4 = 2ab turns into ab = 2 after dividing both sides by 2.

We're told that a,b are rational numbers. Let's assume that they are integers (which is a subset of the rational numbers).

If so, then we have these four possibilities

If a,b are negative, then you'll find that [tex]a\sqrt{2}+b\sqrt{3}[/tex] overall is negative. But this contradicts that [tex]\sqrt{11+4\sqrt{6}}[/tex] is positive. So a,b must be positive.

Let's assume that a = 1 and b = 2. If so, then,

2a^2+3b^2 = 2(1)^2+3(2)^2 = 14

but we want that result to be 11 instead.

Let's try a = 2 and b = 1

2a^2+3b^2 = 2(2)^2+3(1)^2 = 11

which works out perfectly.

Therefore,

[tex]\sqrt{11+4\sqrt{6}} = 2\sqrt{2}+\sqrt{3}\\\\[/tex]

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

Checking the answer:

Use a calculator to find that

[tex]\sqrt{11+4\sqrt{6}} \approx 4.5604779\\\\2\sqrt{2}+\sqrt{3} \approx 4.5604779\\\\[/tex]

both have the same decimal approximation, so this is a fairly informal way to confirm the answer.

Another thing you can do is to take advantage of the idea that if x = y, then x-y = 0

So if you want to see if two things are equal, you subtract them. You should get exactly 0 or something very small (pretty much equal to 0).

Answer:

the answer is 2.8

Step-by-step explanation:

Answer:

[tex]x=4[/tex]

Step-by-step explanation:

Given:

[tex]\frac{3x}{5} -0.5=1.9[/tex]

Add 0.5 to both sides

[tex]\frac{3x}{5} =2.4[/tex]

Multiply 5 from both sides

[tex]3x=12[/tex]

Divide both sides by 3

[tex]x=4[/tex]

Hope this helps

Answer:

3 < x

Step-by-step explanation:

3(8 – 4x) < 6(x – 5)

Divide each side by 3

3/3(8 – 4x) < 6/3(x – 5)

(8 – 4x) < 2(x – 5)

Distribute

8-4x < 2x-10

Add 4x to each side

8-4x+4x < 2x-10+4x

8 < 6x-10

Add 10 to each side

8+10 < 6x-10+10

18 < 6x

Divide by 6

18/6 < 6x/6

3 < x

Answer:

39.52

Step-by-step explanation:

Answer:

Table multiplication:

7.6 times

5.2 =

15.2

38.0

—-

39.52

Given:

Bananas are $2.40 per pound.

To find:

The graph and unit rate on the graph.

Solution:

Let y be the total cost of x pounds of bananas.

Cost of 1 pound of banana = $2.40

Cost of x pound of bananas = $2.40x

So, the required equation for the given situation is:

[tex]y=2.4x[/tex]

The graph of [tex]y=2.4x[/tex] describes the proportional relationship between the two quantities.

The graph of [tex]y=2.4x[/tex] is a line passing through (0, 0) with a slope of 2.4; the slope 2.4 is the unit rate of each pound of bananas.

Therefore, the correct option is (b).

Answer:

a

Step-by-step explanation:

Answer:

Step-by-step explanation:

120 : 30

ivans get £90 more

Answer:

[tex]7,500[/tex]

Step-by-step explanation:

Let's solve this problem step-by-step. The library had 1,500 books in 2011. The ratio of books in 2011 and in 2012 is 1:2. Therefore, let the number of books in 2012 be [tex]x[/tex].

We have the following proportion:

[tex]\frac{1}{1,500}=\frac{2}{x},\\x=2\cdot 1,500=3,000[/tex]

Therefore, there were 3,000 books in 2012. The ratio of books in 2012 and in 2013 is 2:5. Let the number of books in 2013 be [tex]y[/tex].

We have:

[tex]\frac{2}{3,000}=\frac{5}{y},\\2y=5\cdot 3,000,\\2y=15,000\\y=\boxed{7,500}[/tex]

Therefore, there were 7,500 books in 2013.

Answer:

7,500 books

Step-by-step explanation: