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Shtirlitz [24]
3 years ago
11

Which word phrases can be modeled by the expression?

Mathematics
2 answers:
tensa zangetsu [6.8K]3 years ago
6 0
A) (x÷7)-10
B) 7x -10
C) (x-10)÷7
D) (x÷7)-10

A and D are the same
kumpel [21]3 years ago
4 0
I'm not sure but I'll try and figure this out.
You might be interested in
11. A jug contains 2 gallons of water. How many quarts of water does the jug contain? ​
s2008m [1.1K]

Answer:

first change 1 gallon into quarts and that will be

4.8038

so 2 quarts = ? cross multiply ans the answer u will get is 9.6076

5 0
2 years ago
The number x is reduced by 25% and then halved
Maksim231197 [3]
Since it says the number  x , this means x represents a number , reduced by 25 means that this is most likely half of 25 so whats half of 25%?

well 5 * 5 = 25 , so here is your answer  = 5

8 0
3 years ago
How do you find the limit?
coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}

We can cancel x-5.

\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}

Then directly substitute x = 5 in.

\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s <em>too easy</em> but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}

Differentiate numerator and denominator, apply power rules.

<u>Differential</u> (Power Rules)

\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}

<u>Differentiation</u> (Property of Addition/Subtraction)

\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}

Hence from the expressions,

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}

<u>Differential</u> (Constant)

\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \  is \ \ a \ \ constant.)}}

Therefore,

\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}

Now we can substitute x = 5 in.

\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\&#10;&#10;\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}

Thus, the limit value is 2/5 same as the first method.

Notes:

  • If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.
8 0
2 years ago
1. Write a numerical expression for the phrase “twelve groups of the sum of fourteen and 10.”
Rom4ik [11]
The expression for five groups of a number added to seventeen.

Let the number be X
Five groups of the number will be 5X
The required expression will be
5(X) +17
6 0
3 years ago
10. The lengths of the electrical extension cords in a workshop are 6 ft, 8 ft, 25 ft, 8 ft, 12 ft, 50 ft, and 25 ft. What are t
Nadya [2.5K]
6 ft= 6x12in= 72 in.
8 ft= 8x12in= 96 in.
12ft=12x12in=144 in
25ft=25x12in=300 in
50ft=50x12in=600 in

72 in, 96 in, 96 in, 144 in, 300 in, 300 in, 600 in
Mean= (72+96+96+144+300+300+600)÷7
1608÷7 = 229.7 inches
Median=middle value in set = 144 inches
Mode= value(s) tha occur most often = 96 inches and 300 inches
Range=difference of largest and smallest values in set = 600-72= 528
Choice A
3 0
3 years ago
Read 2 more answers
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