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3 years ago

Can anyone help?!!!!!!!

Mathematics

2 answers:

Mamont248 [21]3 years ago

6 0

Since it is on the graph, you know that the factored form gives you the roots. If you look at the graph, the x intercepts are the roots. So x=-2, x=-1, and x=3. That means factored form would be (x+2)(x+1)(x-3)

That’s choice D!

Alexus [3.1K]3 years ago

6 0

The roots of the equation are:x = -2, -1 and +3Therefore, the factored equation should have the opposite signs:(x +2) * (x +1) * (x -3)
The answer is D)

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Zelda completed the right column of the table to help her find the difference of the fractions. 5/6 - 1/9

ArbitrLikvidat [17]

Answer:

13/18

Step-by-step explanation:

1. $\frac{5}{6} -\frac{1}{9}$

2. $\frac{n}{54} - \frac{n}{54}$

3. $\frac{45}{54} - \frac{6}{54}$

4. $= \frac{39}{54}$

5. $Simplify: \frac{13}{18}$

4 0

3 years ago

Identify the equation in slope-intercept form for the line with the given slope that contains the given point. Slope =2; (1,4)

Elina [12.6K]

Slope-intercept form: y=mx+b, with m being the slope and b being the y-intercept. First, plug in the given slope into the slope form:

y=2x+b

Now, given the coordinates (1,4), you can plug the coordinates into the equation for x and y respectively, to solve for b (the y-intercept).

4=2(1)+b

4=2+b

Subtract 2 from both sides.

2=b

The y-intercept is 2.

Now, plug that into the equation:

y=2x+2

The answer is y=2x+2.

I hope this helps :)

3 0

3 years ago

Can someone help me out

kvasek [131]

It equals 5? The markings means it’s equal right?

3 0

3 years ago

Find cos y and tan y if csc y = -√6/2 and cot y >0.

fomenos

Answer:

$\cos y = -\dfrac{\sqrt{3} }{3}$

$\tan y = \sqrt{2}$

Step-by-step explanation:

Recall that

$\boxed{\csc y := \dfrac{1}{\sin y}}$

$\boxed{\cot y := \dfrac{\cos y}{\sin y}}$

We know that

$\csc y = \dfrac{-\sqrt{6} }{2}$

Note that according to the definition of $\csc y$ it is true that both sine and cosine are negative, once $\csc y = \dfrac{-\sqrt{6} }{2}$ . Because $\cot y > 0$ , this conclusion is true. We basically have

$\boxed{(-a)(1/-b)=a/b \text{ such that } a,b\in\mathbb{R}_{\geq 0}}$

Sure it is true $\forall y\in\mathbb{R}$ but perhaps this way is better to understand.

In order to find sine, we can use the definition and manipulate the rational expression.

$\csc y = \dfrac{-\sqrt{6} }{2} = \dfrac{-\sqrt{6} / -\sqrt{6} }{2/-\sqrt{6} } = \dfrac{1 }{-\dfrac{2}{\sqrt{6} } }$

Therefore,

$\sin y =-\dfrac{2}{\sqrt{6} }$

Here I just divided numerator and denominator by $-\sqrt{6}$ .

Now, to find cosine we can use the identity

$\boxed{\sin^2y +\cos ^2y =1}$

Thus,

$\left(-\dfrac{2}{\sqrt{6} }\right)^2 + \cos ^2y =1 \implies \dfrac{4}{6 } +\cos ^2y =1$

$\implies \cos ^2y =1 - \dfrac{4}{6 } \implies \cos ^2y =\dfrac{1}{3 } \implies \cos y = \pm \dfrac{\sqrt{1} }{\sqrt{3} } = \pm \dfrac{\sqrt{1} \sqrt{3} }{3} = \pm \dfrac{\sqrt{3} }{3}$

$\cos y = \pm\dfrac{\sqrt{3} }{3}$

Once we have $\cot y > 0$ , we just consider

$\cos y = -\dfrac{\sqrt{3} }{3}$

FInally, for tangent, just consider

$\boxed{\tan y := \dfrac{\sin y}{\cos y}}$

thus,

$\tan y = \dfrac{\sin y}{\cos y} = \dfrac{-\frac{2}{\sqrt{6} }}{-\frac{\sqrt{3} }{3}} = \dfrac{6}{\sqrt{18} } =\dfrac{6}{3\sqrt{2} } =\dfrac{2}{\sqrt{2} } = \sqrt{2}$

5 0

3 years ago

Hi can you make the numbers

elena-14-01-66 [18.8K]

For 44
7x2=14
14+8=22
22x2=44
Is this ok, I hope I helped :)

4 0

3 years ago