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3 years ago

A segment had a coordinates (-4, 9) and (0, 7). What is the midpoint of the segment

1 answer:

4 0

The x-coordinate of the midpoint is the average of the x-coordinates of the endpoints. The y-coordinate of the midpoint is the average of the y-coordinates of the endpoints.

x-coordinate of midpoint: (-4 + 0)/2 = -4/2 = -2

y-coordinate of midpoint: (9 + 7)/2 = 16/2 = 8

Answer: The midpoint is (-2, 8).

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Plz ASAP <br> I just need help w solving it (explanation) don’t need the answer

lyudmila [28]

x^a / x^b = x^(a-b)

so here its x^( (5/6)- (1/6) )

= x^ 4/6 or ^2/3

6 0

3 years ago

What is the area of a section of pavement that is 20 ft wide and 70 yd long?

myrzilka [38]

Answer:

A = 4200 ft^2

Step-by-step explanation:

We know the formula for area is

A = l*w

We need to have the same units

convert yd to ft

1 yd = 3ft

Multiply each by 70

70 yds = 210 ft

A = 210 *20

A = 4200 ft^2

6 0

3 years ago

What is 9+594,327= <br> I just was wondering?!?!

astraxan [27]

Answer: 594336

Step-by-step explanation: Because of google

4 0

2 years ago

Read 2 more answers

Please help with any of this Im stuck and having trouble with pre calc is it basic triogmetric identities using quotient and rec

german

How I was taught all of these problems is in terms of r, x, and y. Where sin = y/r, cos = x/r, tan = y/x, csc = r/y, sec = r/x, cot = x/y. That is how I will designate all of the specific pieces in each problem.

Let's start with sin here. $\frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}}$ Therefore, because sin is y/r, r = $\sqrt{5}$ and y = +2. Moving over to cot, which is x/y, x = -1, and y = 2. We know y has to be positive because it is positive in our given value of sin. Now, to find cos, we have to do x/r.

cos = $\frac{-1}{\sqrt{5}} = \frac{-\sqrt{5}}{5}$

Let's start with secant here. Secant is r/x, where r (the length value/hypotenuse) cannot be negative. So, r = 9 and x = -7. Moving over to tan, x must still equal -7, and y = $4\sqrt{2}$ . Now, to find csc, we have to do r/y.

csc = $\frac{9}{4\sqrt{2}} = \frac{9\sqrt{2}}{8}$

The pythagorean identities are

sin^2 + cos^2 = 1,

1 + cot^2 = csc^2,

tan^2 + 1 = sec^2.

Let's take a look at the information given here. We know that cos = -3/4, and sin (the y value), must be greater than 0. To find sin, we can use the first pythagorean identity.

sin^2 + (-3/4)^2 = 1

sin^2 + 9/16 = 1

sin^2 = 7/16

sin = $\sqrt{7/16}$ = $\frac{\sqrt{7}}{4}$

Now to find tan using a pythagorean identity, we'll first need to find sec. sec is the inverse/reciprocal of cos, so therefore sec = -4/3. Now, we can use the third trigonometric identity to find tan, just as we did for sin. And, since we know that our y value is positive, and our x value is negative, tan will be negative.

tan^2 + 1 = (-4/3)^2

tan^2 + 1 = 16/9

tan^2 = 7/9

tan = - $\sqrt{7/9} = \frac{-\sqrt{7}}{3}$

Let's take a look at the information given here. If we know that csc is negative, then our y value must also be negative (r will never be negative). So, if cot must be positive, then our x value must also be negative (a negative divided by a negative makes a positive). Let's use the second pythagorean identity to solve for cot.

1 + cot^2 = $(\frac{-\sqrt{6}}{2})^{2}$

1 + cot^2 = 6/4

cot^2 = 2/4

cot = $\frac{\sqrt{2}}{2}$

tan = $\sqrt{2}$

Next, we can use the third trigonometric identity to solve for sec. Remember that we can get tan from cot, and cos from sec. And, from what we determined in the beginning, sec/cos will be negative.

$(\frac{2}{\sqrt{2}})^2$ + 1 = sec^2

4/2 + 1 = sec^2

2 + 1 = sec^2

sec^2 = 3

sec = - $\sqrt{3}$

cos = $\frac{-\sqrt{3}}{3}$

Hope this helps!! :)

3 0

2 years ago

Read 2 more answers

Where does the point (0, - 3) lie?

gregori [183]

An ordered pair is written like this, ( x, y ). In this case x = 0 and y = -3. On a graph the vertical line is the y-axis and the horizontal line is the x-axis. The origin is point ( 0, 0 ). To the left of the origin on the x-axis is the negative number line and to the right is the positive number line. On the y-axis, south of the origin is the negative number line and north is the positive number line. When you plot a point on a graph you do x first, so if x equals 1, you would move one right, -1, one left. IF y were to equal 2 then from the place where you are on the x-axis, 1, you would move two up, -2, two down. In this case x = 0 so you would stay at the origin, and y = -3 so you would move 3 down. So ( 0, -3 ) would lie negative y-axis. The answer is D.

3 0

2 years ago