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3 years ago

It says it in the image please help me

Mathematics

1 answer:

r-ruslan [8.4K]3 years ago

6 0

Answer:

The answer you have selected should be correct.

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3x^2-12x-21=0 solve by completing the square

morpeh [17]

Answer:

x = 2 ± √11 (I believe)

Step-by-step explanation:

3 0

3 years ago

Which is an equation of the line with slope –3 and passes through (2, –1)

Savatey [412]

using the formula:

y = mx +b

Using the given slope and coordinate points we get:

-1 = -3(2) + b

-1 = -6 + b

add 6 to each side

5 = b

y = -3x + 5

which is the same as:

3x + y = 5 ...answer choice a

7 0

2 years ago

Which of the following is not a Pythagorean triple?

zavuch27 [327]

Answer:

Step-by-step explanation:

4 0

3 years ago

F(x) =2x^2+12x-6 Does this function have a minimum or maximum value? What is this minimum or maximum value?

Alex777 [14]

Let's compare the given function with the model for a quadratic equation:

$\begin{gathered} f(x)=ax^2+bx+b \\ a=2,b=12,c=-6 \end{gathered}$

Since the value of a is positive, the parabola has its concavity upwards, and the function has a minimum value.

The minimum value can be found calculating the y-coordinate of the vertex:

$\begin{gathered} x_v=-\frac{b}{2a}=-\frac{12}{4}=-3 \\ \\ y_v=2\cdot(-3)^2+12\cdot(-3)-6 \\ y_v=2\cdot9-36-6^{} \\ y_v=-24 \end{gathered}$

Therefore the minimum value is -24.

4 0

1 year ago

If x= sin theta then x/√1-x^2 is

SCORPION-xisa [38]

Well,

Given that $x=\sin(\theta)$ ,

We can rewrite the equation like,

$\dfrac{\sin(\theta)}{\sqrt{1-\sin(\theta)^2}}$

Now use, $\cos(\theta)^2+\sin(\theta)^2=1$ which implies that $1-\sin(\theta)^2=\cos(\theta)^2$

That means that,

$\dfrac{\sin(\theta)}{\sqrt{1-\sin(\theta)^2}}\Longleftrightarrow\dfrac{\sin(\theta)}{\sqrt{\cos(\theta)^2}}$

By def $\sqrt{x^2}=x$ therefore $\sqrt{\cos(\theta)^2}=\cos(\theta)$

So the fraction now looks like,

$\dfrac{\sin(\theta)}{\cos(\theta)}$

Which is equal to the identity,

$\boxed{\tan(\theta)}=\dfrac{\sin(\theta)}{\cos(\theta)}$

Hope this helps.

r3t40

6 0

3 years ago