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Leya [2.2K]
3 years ago
15

Find the measure of angle x in the figure below:

Mathematics
2 answers:
zhuklara [117]3 years ago
4 0

Answer:

70°

Step-by-step explanation:

55 plus 55 = 110

180 - 110 = 70

geniusboy [140]3 years ago
4 0
70° because the triangle must equal 180°. Therefore, 55+55=110. If the triangle must equal 180° then you subtract by the 2 angles that you have from 180°. That should get you to 180°-110° which equals 70°
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-25 - 30 + 2 = pls help
Galina-37 [17]

Answer:

7

Step-by-step explanation:

Do addition first, so 30 plus 2, and then subtract, 32 minus 25. It is seven.

5 0
3 years ago
Ashot borrowed $225, and paid $108 as an interest of eight years. What was the annual interest rate?
Helga [31]

Answer:

6%

Step-by-step explanation:

I asked this same exact question, no one got the right answer, so I just guessed, and what do you know, I got the answer right! So here you go.

7 0
3 years ago
Read 2 more answers
What is the volume of the rectangular prism below in cubic inches?
lions [1.4K]

Answer:

24

Step-by-step explanation:

The volume formula is all the dimensions multiplied together, or lengthxwidthxheight.

So, you do 1x1x2 which is 2 feet. HOWEVER, it is looking for CUBIC INCHES, not cubic FEET.

There are twelve inches in a foot so 2x12=24 inches.

5 0
2 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
3 years ago
Is (5, 10), (-8, 3), (-1, 7), (-5, 10) A Function ?
ankoles [38]

Answer:

Step-by-step explanation:

Notice that the inputs {5, -8, -1, -5} are all different.  So we can immediately conclude that this is a function.

If the inputs were {5, -8, -1, 5}, this would not be a function, as the same input value (5) would result in two different outputs.

6 0
3 years ago
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