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Eddi Din [679]
3 years ago
14

If f(x) = 2x^2+ 1 and g(x)= x^2-7, find (f-g)(x)

Mathematics
1 answer:
Ganezh [65]3 years ago
4 0

Answer:

2x squared +1 (-)x squared -7 is

a.) x squared +8

Step-by-step explanation:

:)

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A rectangle's width is 1/3 its length and its perimeter is 40 m. find the dimensions of the rectangle
Rina8888 [55]

Answer:

Length = 15m

Width = 5m

Step-by-step explanation:

w = 1/3l

p = 2l + 2/3l

40 = 2 2/3 l

l = 15

w = 1/3(15)

w = 5

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

8 0
3 years ago
Determine whether the equation is exponential growth or exponential decay
Harman [31]

9514 1404 393

Answer:

  1. decay
  2. decay
  3. growth

Step-by-step explanation:

If the base of a positive exponential is greater than 1, the function is a growth function. If it is less than 1, the function decays.

Remember that a^-1 = 1/a, so a negative exponential can be transformed to a positive one.

__

1. y = 2(11/5)^-x = 2(5/11)^x . . . . 5/11 < 1, so decay

2. y = e^(-2x) = (1/e^2)^x . . . . 1/e^2 < 1, so decay

3. y = 1/4e^x . . . . e > 1, so growth

_____

e ≈ 2.71828 . . . an irrational number

4 0
3 years ago
Please help on this math question!!!
Andre45 [30]
I think it’s gonna be 6 adults for every 42 students not really sure so sorry if I got it wrong
6 0
2 years ago
Evaluate the expression. 2x – 19, when x = –11 3 –41 –3 41
kakasveta [241]
Substitute (-11) in for x. You will get 2(-11)-19. Then, use the order of operations to get -22-19=
-41.
Hope that helps!
3 0
3 years ago
Find the limit, if it exists, or type dne if it does not exist.
Phantasy [73]
\displaystyle\lim_{(x,y)\to(0,0)}\frac{\left(x+23y)^2}{x^2+529y^2}

Suppose we choose a path along the x-axis, so that y=0:

\displaystyle\lim_{x\to0}\frac{x^2}{x^2}=\lim_{x\to0}1=1

On the other hand, let's consider an arbitrary line through the origin, y=kx:

\displaystyle\lim_{x\to0}\frac{(x+23kx)^2}{x^2+529(kx)^2}=\lim_{x\to0}\frac{(23k+1)^2x^2}{(529k^2+1)x^2}=\lim_{x\to0}\frac{(23k+1)^2}{529k^2+1}=\dfrac{(23k+1)^2}{529k^2+1}

The value of the limit then depends on k, which means the limit is not the same across all possible paths toward the origin, and so the limit does not exist.
8 0
4 years ago
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