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Dovator [93]
3 years ago
12

Help please algebra

Mathematics
1 answer:
Murljashka [212]3 years ago
4 0

The answer is D. -24 - 49i/13

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Domain and range of linear and quadratic functions worksheet
Vlad [161]

If the value of a is negative, then the range will be (-∞, k) and if the value of the a is positive then the range will be (k, ∞).

<h3>What is a quadratic equation?</h3>

It's a polynomial with a worth of nothing.

There exist polynomials of variable power 2, 1, and 0 terms.

A quadratic condition is a condition with one explanation where the degree of the equation is 2.

Domain and range of linear and quadratic functions

Let the linear equation be y = mx + c.

Then the domain and the range of the linear function are always real.

Let the quadratic equation will be in vertex form.

y = a(x - h)² + k

Then the domain of the quadratic function will be real.

If the value of a is negative, then the range will be (-∞, k) and if the value of the a is positive then the range will be (k, ∞).

More about the quadratic equation link is given below.

brainly.com/question/2263981

#SPJ4

7 0
1 year ago
Write the equation using function notation f(x). then find f(0)<br><br> 2x^2 - 3y = 6
Grace [21]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \: f(0) = -2

____________________________________

\large \tt Solution  \: :

\qquad \tt \rightarrow \: 2 {x}^{2}  - 3y = 6

\qquad \tt \rightarrow \: 3y + 6 = 2 {x}^{2}

\qquad \tt \rightarrow \: 3y = 2 {x}^{2}  - 6

\qquad \tt \rightarrow \: y =  \cfrac{2 {x}^{2} - 6 }{3}

\qquad \tt \rightarrow \: y =  \cfrac{2 {}^{} }{3}  {x}^{2}  - 2

[ here, y can be replaced with f(x) because y is a function of x ]

\qquad \tt \rightarrow \: f(x) =  \cfrac{2 {}^{} }{3}  {x}^{2}  - 2

\large\textsf{Find f(0):}

\qquad \tt \rightarrow \: f(0) =  \cfrac{2 {}^{} }{3}  {(0)}^{2}  - 2

\qquad \tt \rightarrow \: f(0) =  0  - 2

\qquad \tt \rightarrow \: f(0) =   - 2

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

8 0
2 years ago
Functions &amp; (f o g) (#)
sertanlavr [38]
Hello there!

I'm assuming since there is no question, that you want an explanation for composite functions.

Today, I want to introduce you to a very new way of looking at functions. Think of a function as a machine. I'll call this machine f. When you plug something into this machine, it is an x-value. The machine changes the x-value into a new value which is called a y-value. This is how a function works.

With composite functions though, things get a little bit tricky. To make f(g(x)), you need to plug in x into the g machine, and it will give you an output. (y-value) The next thing you do is take that y-value and plug it into the g machine. The g machine then gives you a new value. This value is f(g(x)).

Let's do an example together...
f(x)=3x and g(x)=x²+4
if we want f(g(x)), first plug in x to the g machine. when plugging in x to the g machine, we get x²+4 as given in the question.
Now we must plug in g(x) into the f machine. Since g(x) is x²+4, we just replace x with x²+4.
We get 3(x²+4)
This means that f(g(x))=3(x²+4)

NOTE: If you are seeking help with an actual question, please message me in the comments and I will assist you shortly!

I hope this helps!
Best wishes :)
7 0
3 years ago
My sister is half my age when I was 6 now im 70 how old is she?​
TEA [102]

Answer:

your sister would be 35

Step-by-step explanation:

5 0
2 years ago
Find the limit. use l'hospital's rule if appropriate. if there is a more elementary method, consider using it. lim x→1 1 − x + l
AfilCa [17]

\displaystyle\lim_{x\to1}\frac{1-x+\ln x}{1+\cos5\pi x}

Evaluating the limand directly at x=1 gives an indeterminate form \dfrac00. Apply L'Hospital's rule once and we get

\displaystyle\lim_{x\to1}\frac{-1+\frac1x}{-5\pi\sin5\pi x}

Again, plugging in x=1 returns \dfrac00. Apply the rule once more:

\displaystyle\lim_{x\to1}\frac{-\frac1{x^2}}{-25\pi^2\cos5\pi x}=\frac1{25\pi^2}\lim_{x\to1}\frac1{x^2\cos5\pi x}

Now, in the denominator, when x=1 we get x^2\cos5\pi x=-1, so the limit is -\dfrac1{25\pi^2}.

3 0
3 years ago
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