3 years ago

What are the zeros of the function shown in the graph? −1, 1, 2 −2, −1, 1 −3, −1, 1 −1, 1, 3

Mathematics

1 answer:

joja [24]3 years ago

8 0

Answer:

Step-by-step explanation:

You might be interested in

A color printer prints 25 pages in 10 minutes. How many pages does it print per minute

alexdok [17]

25 (pgs) / 10 (min) = 2.5 pgs per minute

8 0

3 years ago

Read 2 more answers

Does someone know the answer of (-9+v)/8=3

vredina [299]

(-9+V) /8 = 3

-9/8 +V/8 = 24/8

-9+V = 24

V = 24+9

V = 33

let's check : -9+33 = 24 and 24/8 = 3

3 0

4 years ago

Jimmy is trying to factor the quadratic equation $ax^2 + bx + c = 0.$ He assumes that it will factor in the form \[ax^2 + bx + c

Shkiper50 [21]

$ax^2+bx+c=(Ax+B)(Cx+D)$

then

$ax^2+bx+c=ACx^2+(AD+BC)x+BD$

$\implies a=AC$

If $a$ is known and Jimmy wants to find $A$ , then he has to know the value of $C$ .

6 0

3 years ago

Read 2 more answers

Use a graph in a (-2π, 2π, π/2) by (-3, 3, 1) viewing rectangle to complete the identity.

yaroslaw [1]

First, notice that:

$2\tan (\frac{x}{2})=2\cdot(\pm\sqrt[]{\frac{1-cosx}{1+\cos x})}$

And in the denominator we have:

$1+\tan ^2(\frac{x}{2})=1+\frac{1-\cos x}{1+\cos x}=\frac{1+cosx+1-\cos x}{1+cosx}=\frac{2}{1+\cos x}$

then, we have on the original expression:

$\begin{gathered} \frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}=\frac{2\cdot\pm\sqrt[]{\frac{1-\cos x}{1+cosx}}}{\frac{2}{1+\cos x}}=\frac{2\cdot(\pm\sqrt[]{1-cosx})\cdot(1+\cos x)}{2\cdot(\sqrt[]{1+cosx})} \\ =(\sqrt[]{1-\cos x})\cdot(\sqrt[]{1+\cos x})=\sqrt[]{(1-\cos x)(1+\cos x)} \\ =\sqrt[]{1-\cos^2x}=\sqrt[]{\sin^2x}=\sin x \end{gathered}$

therefore, the identity equals to sinx

8 0

1 year ago

True or false ? the binary number 00110010 represents a byte

lina2011 [118]

Answer:

true

Step-by-step explanation:

do you play FN? I want to make a friend

3 0

3 years ago