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soldier1979 [14.2K]
3 years ago
5

Angles that share a vertex, but not a side are called?

Mathematics
1 answer:
WARRIOR [948]3 years ago
5 0
Adjacent angles are the angles that share a common vertex but no common interior points.
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What dose the hypotenuse=?
Kitty [74]
Hypotenuse by Pythagoras theorem = 407.83.
3 0
3 years ago
4x^2-2x-4=0 solve using quadratic formula
MArishka [77]

Answer:

Step-by-step explanation:

Although you can start with 4x^2-2x-4=0, it's just slightly easier to reduce the coefficients.  4x^2-2x-4=0 becomes 2x^2 - x - 2 = 0.

The discriminant (part of the quadratic formula) is b^2 - 4ac.  Using b = -1. a = 1 and c = -2, we get (-1)^2 - 4(2)(-2), or 17.

                                                       -b ± √(b^2 - 4ac)

Then the quadratic formula   x = ---------------------------

                                                                 2a

                                         -(-1) ± √17           1 ± √17

takes on the values  x = -----------------  =  ---------------

                                                  4                      4

4 0
3 years ago
Find the slope of a line whose equation is 3X+6Y=9
Lisa [10]

To find the slope of the above equation, it is easiest to put it into slope-intercept form, y=mx + b, where the variable m represents the slope. To do this, we must isolate the variable y on the left side of the equation by using the reverse order of operations. First, we should subtract 3x from both sides of the equation.

3x + 6y = 9

6y = -3x + 9

Next, we should divide both sides of the equation by 6 to undo the coefficent of 6 on the variable y.

y = -1/2x + 3/2

Therefore, the slope of the line is -1/2 (the coefficient of the variable x in slope-intercept form).

Hope this helps!

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
3 years ago
Plz help I suck at math and need help on this.... divided by 18
MatroZZZ [7]
2 divided by 18 is 9 and 3 divided by 18 is 6 hope that helps
6 0
3 years ago
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