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Y_Kistochka [10]

3 years ago

12

Which methods could you use to calculate the y-coordinate of the midpoint of a vertical line segment with endpoints at (0,0) and

(0, -12)?

1 answer:

Olenka [21]3 years ago

7 0

Answer:

add the y coordinates of the two lines and divide by 2

Step-by-step explanation:

½[0+(-12)]

½[0-12]

½[-12)

y co-ordinate of the midpoint is -6

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2 years ago

A total solar eclipse in 2003 lasted 8 9/20 minutes and was viewed from the southern hemisphere. The next total solar eclipse in

beks73 [17]

Answer:

5/101/180

Step-by-step explanation:

6 0

2 years ago

Select the set of equations that would be used to solve the following absolute value equation:|x-5|=20

-BARSIC- [3]

X-5= 20
x-5= -20

Absolute value makes a negative positive, so when you are solving x-5 could have been 20 originally or -20 originally. Solve for both.

7 0

3 years ago

Read 2 more answers

Given the quadratic equation <img src="https://tex.z-dn.net/?f=y%20%3D%202%28x%20-1%29%5E%7B2%7D%20%2B%208" id="TexFormula1" tit

Sunny_sXe [5.5K]

Answer:

Part 1) "a" value is $2$

Part 2) The vertex is the point $(1,8)$

Part 3) The equation of the axis of symmetry is $x=1$

Part 4) The vertex is a minimum

Part 5) The quadratic equation in standard form is $y=2x^{2}-4x+10$

Step-by-step explanation:

we know that

The equation of a vertical parabola into vertex form is equal to

$y=a(x-h)^{2}+k$

where

(h,k) is the vertex of the parabola

if a > 0 then the parabola open upward (vertex is a minimum)

if a < 0 then the parabola open downward (vertex is a maximum)

The equation of the axis of symmetry of a vertical parabola is equal to the x-coordinate of the vertex

so

$x=h$

In this problem we have

$y=2(x-1)^{2}+8$ -----> this is the equation in vertex form of a vertical parabola

The value of $a=2$

so

a>0 then the parabola open upward (vertex is a minimum)

The vertex is the point $(1,8)$

so

$(h,k)=(1,8)$

The equation of the axis of symmetry is $x=1$

The equation of a vertical parabola in standard form is equal to

$y=ax^{2}+bx+c$

Convert vertex form in standard form

$y=2(x-1)^{2}+8$

$y=2(x^{2}-2x+1)+8$

$y=2x^{2}-4x+2+8$

$y=2x^{2}-4x+10$

see the attached figure to better understand the problem

7 0

3 years ago

Find cos q if q is an angle in standard position and the point with coordinates (3, –4) lies on the terminal side of the angle.

gayaneshka [121]

Answer:

cos q = 3/5

Step-by-step explanation:

Standard position means the vertex (point or corner of the angle) is at (0,0) and one side of the angle is glued to the positive x-axis (facts, but not technical math terms) See image. Special triangles have all three sides nice and clean with whole number lengths, we call these Pythagorean triples. 3-4-5 is your most basic Pythagorean triple. So we don't even have to calculate the hypotenuse, see image. Now the triangle shown is easy to work with, using entry-level trig...cos = ADJ/HYP. So we get 3/5=.6 BUUuuuut, the angle q in the original problem is actually the giant angle, marked in yellow (see image) and we're in the fourth quadrant which means there's negative numbers all over the place. So just to be sure the answer is .6 and not -.6 Check your signs. One trick to remember is the ASTC markings in the quadrants. I use All Students Take Calculus, but what it means is in the first quadrant All the trig functions are positive. Only Sine (and fam) are positive in the 2nd quadrant. Tan (and fam) in the 3rd and Cos and fam in the 4th quadrant. It's a good quick check.

cos q = 3/5 OR cos q = .6

8 0

2 years ago