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4 years ago

10

Find the midpoint of the segment with the following endpoints. (-1,3) and (-9, -7)

2 answers:

Pachacha [2.7K]4 years ago

7 0

The midpoint for theses should be (-5,-2)

Mrrafil [7]4 years ago

5 0

(-5,-2)
Average the x and y values

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XYZ Music Store sells new CDs,n, for $ 13.50, and used $6.50 .The store eamed $ 456 revenue last month. The store sold 14 more u

marysya [2.9K]

Answer:

c

Step-by-step explanation:

i did it in a test and i used a calculator

4 0

3 years ago

A particle moves along the x-axis so that at time t its position is given by s(t) = (t + 1)(t - 3)3, t > 0. /p>For what va

Zepler [3.9K]

Given the position function, the velocity function is obtained by taking the derivative:

$s(t)=(t+1)(t-3)^3\implies v(t)=(t-3)^3+3(t+1)(t-3)^2$

The velocity is increasing its own derivative is positive, so we also have to find the acceleration by taking another derivative:

$a(t)=4(t-3)^3+3(t-3)^2+6(t+1)(t-3)$

To find when $a(t)>0$ , we first need to know where $a(t)=0$ :

$4(t-3)^3+3(t-3)^2+6(t+1)(t-3)=(t-3)\bigg(4(t-3)^2+3(t-3)+6(t+1)\bigg)=(t-3)(4t^2-15t+33)=0$

The quadratic factor is always positive (its discriminant is negative), which leaves one solution at $t=3$ . To either side of $t=3$ , we have, for instance,

$a(2)=-12$

$a(4)=36>0$

which indicates that $v(t)$ is increasing for $t>3$ , making the answer A.

5 0

3 years ago

10:100 it equivalent to

wlad13 [49]

1:10 is one thing that is equivalent

7 0

3 years ago

Read 2 more answers

Can someone please answer this equation?

Airida [17]

Answer:

4

Step-by-step explanation:

4

7 0

4 years ago

If tan theta= 3/4 , find csc theta

dangina [55]

Answer:

$csc\theta=\frac{5}{3}$

Step-by-step explanation:

Given:

$tan\theta =\frac{3}{4}$

$cot\theta =\frac{4}{3}$ [∵ $cot\theta =\frac{1}{tan\theta}$ ]

Squaring both sides.

$cot^2\theta =\frac{4^2}{3^2}=\frac{16}{9}$

$csc^2\theta -1=\frac{16}{9}$ [∵ $cot^2\theta =csc^2\theta-1]$ ]

Adding 1 to both sides.

$csc^2\theta -1+1=\frac{16}{9}+1$

$csc^2\theta =\frac{16}{9}+1$

$csc^2\theta=\frac{16}{9} +\frac{9}{9}$ [Taking LCD=9 and adding fractions ]

$csc^2\theta=\frac{16+9}{9}$

$csc^2\theta=\frac{25}{9}$

Taking square root both sides.

$\sqrt{csc^2\theta}=\sqrt{\frac{25}{9}}$

∴ $csc\theta=\frac{5}{3}$

6 0

3 years ago