All categories

3 years ago

Which graph represents the function below?

Mathematics

1 answer:

Ierofanga [76]3 years ago

4 0

I think its the bottom left, i'm not completely sure though.

You might be interested in

An equation and the first step in its solution are shown: Equation: x2 + 8x - 9 = 0

Novay_Z [31]

Answer- A-16

Because when doing completing the square, you divide the coefficient of x by 2, which is 4, and the square it which is 16

3 0

2 years ago

Can someone help with this ?

erik [133]

Answer: b=-12

Step-by-step explanation:

120/-10 = -12

-10b/-10 = b

b=-12

3 0

3 years ago

Read 2 more answers

If , find the value of 2 square root of 17-x = 2x-10, find the value of x/4.

ira [324]

$2 \sqrt{17-x} =2x-10\ \ \ \ \Rightarrow\ \ \ D:(17-x \geq 0\ \ \ and\ \ \ 2x-10 \geq 0)\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \leq 17\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \geq 5\\\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ D=\\\\2 \sqrt{17-x} =2(x-5)\ /:2\\\\ \sqrt{17-x} =x-5\ \ \ \Rightarrow\ \ \ (\sqrt{17-x})^2 =(x-5)^2\\\\17-x=x^2-10x+25\\\\$

$x^2-9x+18=0\\\\ x^2-3x-6x+18=0\\\\x(x-3)-6(x-3)=0\\\\(x-3)(x-6)=0\ \ \ \ \Leftrightarrow\ \ \ (x-3=0\ \ \ \ or\ \ \ \ x-6=0)\\\\x=3\ \notin\ D\ \ \ \ \ \ \ \ \ \ \ x=6\ \in\ D\ \ \ \Rightarrow\ \ \ \frac{x}{4} = \frac{6}{4} =1.5\\\\Ans.\ \frac{x}{4} =1.5$

7 0

3 years ago

Show tan(???? − ????) = tan(????)−tan(????) / 1+tan(????) tan(????)<br> .

anyanavicka [17]

Answer:

See the proof below

Step-by-step explanation:

For this case we need to proof the following identity:

$tan(x-y) = \frac{tan(x) -tan(y)}{1+ tan(x) tan(y)}$

We need to begin with the definition of tangent:

$tan (x) =\frac{sin(x)}{cos(x)}$

So we can replace into our formula and we got:

$tan(x-y) = \frac{sin(x-y)}{cos(x-y)}$ (1)

We have the following identities useful for this case:

$sin(a-b) = sin(a) cos(b) - sin(b) cos(a)$

$cos(a-b) = cos(a) cos(b) + sin (a) sin(b)$

If we apply the identities into our equation (1) we got:

$tan(x-y) = \frac{sin(x) cos(y) - sin(y) cos(x)}{sin(x) sin(y) + cos(x) cos(y)}$ (2)

Now we can divide the numerator and denominato from expression (2) by $\frac{1}{cos(x) cos(y)}$ and we got this:

$tan(x-y) = \frac{\frac{sin(x) cos(y)}{cos(x) cos(y)} - \frac{sin(y) cos(x)}{cos(x) cos(y)}}{\frac{sin(x) sin(y)}{cos(x) cos(y)} +\frac{cos(x) cos(y)}{cos(x) cos(y)}}$

And simplifying we got:

$tan(x-y) = \frac{tan(x) -tan(y)}{1+ tan(x) tan(y)}$

And this identity is satisfied for all:

$(x-y) \neq \frac{\pi}{2} +n\pi$

8 0

3 years ago

I need help!!

hammer [34]

Answer:The "mean" is the "average" you're used to, where you add up all the numbers and then divide by the number of numbers. The "median" is the "middle" value in the list of numbers.

Step-by-step explanation:

4 0

3 years ago