6) HELP I got it wrong can someone plz fix it and tell me what I did wrong!

Mathematics

2 answers:

Marianna [84]3 years ago

8 0

The slope is wrong. You go up 6 and over 3 so there for 6/3 which reduces to 2 if you count from the Y point up to the X point you get 6

grigory [225]3 years ago

5 0

Answer:hejdj

Step-by-step explanation:

Hdndjdjd

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I really need help with this question

Veronika [31]

G(x) appears to be a vertical expansion of f(x) by a factor of 3. f(x) appears to be
.. f(x) = x²
so g(x) would be
.. g(x) = 3x²

6 0

3 years ago

Factor completely: 2x2 + 10x + 12

Bess [88]

$\bf 2x^2+10x+12\implies 2(x^2+5x+6)\implies 2(x+3)(x+2)$

recall that 3 * 2 = 6, and 3x + 2x = 5x.

4 0

3 years ago

Read 2 more answers

True/False. Every member of a food web is the prey of another member of the food web.

ra1l [238]

true

Producers --> Consumers --> Primary consumers --> Secondary consumers -->

Tertiary consumers --> Quaternary consumers --> Decomposers --> Producers

and the cycle keeps repeating

<em>Hope it helps... </em>

7 0

3 years ago

Verify the following trigonometric identity: (csc x + cot x)^2 = cos x +1/ 1- cos x

Airida [17]

9514 1404 393

Explanation:

As written, the equation is not an identity. Perhaps you want to show ...

(csc(x) +cot(x))² = (cos(x) +1)/(1 -cos(x))

We will transform the left-side expression to the form of the right-side expression.

$(\csc(x)+\cot(x))^2=\left(\dfrac{1}{\sin(x)}+\dfrac{\cos(x)}{\sin(x)}\right)^2=\dfrac{(1+\cos(x))^2}{\sin(x)^2}\\\\=\dfrac{(1+\cos(x))^2}{1-\cos(x)^2}=\dfrac{1+\cos(x)}{1-\cos(x)}\cdot\dfrac{1+\cos(x)}{1+\cos(x)}=\boxed{\dfrac{\cos(x)+1}{1-\cos(x)}}$