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4 years ago

I need help with this question please.

Mathematics

1 answer:

Rina8888 [55]4 years ago

5 0

I think its -6 because 2(-1)^2 = 2 and + 8(-1) = -6

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Can someone answer my questions please ._.

tekilochka [14]

Answer:

wheres the question?

Step-by-step explanation:

6 0

3 years ago

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50 points!!! Which equation is represented on the graph?

gavmur [86]

Answer:

I think the answer is b ill look into it a Lil more

4 0

3 years ago

The focus of a parabola is (4,7) and the directrix is y=8.

disa [49]

We know that the x-ordinate of the vertex is x = 4, because the x-ordinate of the vertex will not change.

Now, the definition of a parabola in locus terms is that for any fixed point that lies on the parabolic shape, it is equidistant from the directrix to the fixed point and the focus to the fixed point.

Thus, we can deduce that:
PS = PD, where P(x, y) is a moving point that lies on the parabola.

Now, we know that the latus rectum will be -4a, because the directrix is higher than the y-ordinate of the focus. Thus, we can say that the vertex will be the midpoint of the two points y = 7 and y = 8, because that is the only place where the definition of a parabola works.

Hence, we can say that the vertex lies at (4, 7.5).

Thus, our general form becomes:
(x - 4)² = 4a(y - 7.5)

To find a, we can say: 7 = 7.5 + a and 8 = 7.5 - a
a = -0.5, because a is the factor in which we move to find the directrix and focus.

Thus, our equation becomes: (x - 4)² = -2(y - 7.5)

3 0

3 years ago

Find the mean, variance, and standard deviation of the sample with

azamat

Mean:-

$\\ \tt\hookrightarrow \dfrac{\sum x}{n}=\dfrac{210}{10}=21$

Standard Deviation:-

$\\ \tt\hookrightarrow \sqrt{\dfrac{\sum x^2}{n}-(\sum x/n)^2}$

$\\ \tt\hookrightarrow \sqrt{\dfrac{5006}{10}-(21)^2}$

$\\ \tt\hookrightarrow \sqrt{500.6-441}$

$\\ \tt\hookrightarrow \sqrt{59.6}$

$\\ \tt\hookrightarrow 7.72$

Variance:-

$\\ \tt\hookrightarrow (S. D)^2=(7.72)^2=59.6$

6 0

3 years ago

Which of the following points lies on the circle whose center is at the origin and whose radius is 5?

ad-work [718]

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Equation of the circumference whose center is at $\mathsf{C(0,\,0)}$ and whose radius is $\mathsf{r=5:}$

$\mathsf{(x-x_C)^2+(y-y_C)^2=r^2}\\\\
\mathsf{(x-0)^2+(y-0)^2=5^2}\\\\
\mathsf{x^2+y^2=25}$

We are looking for the point that satisfies the equation above.

 • Testing $\mathsf{(-3,\,4):}$

$\mathsf{(-3)^2+4^2}\\\\
\mathsf{9+16}\\\\
\mathsf{25\qquad\quad\checkmark}$

The point $\mathsf{(-3,\,4)}$ lies in the circumference.

 • Testing $\mathsf{(1,\,-2):}$

$\mathsf{1^2+(-2)^2}\\\\ \mathsf{1+4}\\\\ \mathsf{5\ne 25\qquad\quad\diagup\hspace{-9}\diagdown}$

The point $\mathsf{(1,\,-2)}$ doesn't lie in the circumference.

 • Testing $\mathsf{(\sqrt{5},\,\sqrt{5}):}$

$\mathsf{(\sqrt{5})^2+(\sqrt{5})^2}\\\\ \mathsf{5+5}\\\\ \mathsf{10\ne 25\qquad\quad\diagup\hspace{-9}\diagdown}$

So, the point $\mathsf{(\sqrt{5},\,\sqrt{5})}$ doesn't lie in the circumference either.

Answer: (–3, 4).

I hope this helps. =)

8 0

4 years ago

Read 2 more answers