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2 years ago

PLZ HELP NNOOOOOOOWW!! 15 points!!

1 answer:

4 0

The midpoint of the line having the endpoints $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_{1}+_x_{2}}{2}, \frac{y_{1}+_y_{2}}{2})$
basically average them

so given that the line has the endpoints of (-3,7) and (9,-2)
x₁=-3
y₁=7
x₂=9
y₂=-2

so the midpoint can be found by doing
$(\frac{-3+9}{2}, \frac{7+(-2)}{2})=$
$(\frac{6}{2}, \frac{7-2}{2})=$
$(3, \frac{5}{2})$

the midpoint is $(3,\frac{5}{2})$

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How many real number solutions does the equation have 0=5x^2+2x-12

Harman [31]

Rearrange the equation to standard form of a quadratic equation (ax^2+bx+c=0) by switching sides: 5x^2+2x-12=0. Now, use the quadratic equation formula to solve. You should come out with x_1=sqrt61-1/5 and x_2=-1+sqrt61/5. Thus, your answer is B, or two solutions.

7 0

3 years ago

Read 2 more answers

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station. Find the ra

Anika [276]

Answer:

The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

Step-by-step explanation:

Given information:

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station.

$z=1$

$\frac{dx}{dt}=430$

We need to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

$y=2$

According to Pythagoras

$hypotenuse^2=base^2+perpendicular^2$

$y^2=x^2+1^2$

$y^2=x^2+1$ .... (1)

Put z=1 and y=2, to find the value of x.

$2^2=x^2+1^2$

$4=x^2+1$

$4-1=x^2$

$3=x^2$

Taking square root both sides.

$\sqrt{3}=x$

Differentiate equation (1) with respect to t.

$2y\frac{dy}{dt}=2x\frac{dx}{dt}+0$

Divide both sides by 2.

$y\frac{dy}{dt}=x\frac{dx}{dt}$

Put $x=\sqrt{3}$ , y=2, $\frac{dx}{dt}=430$ in the above equation.

$2\frac{dy}{dt}=\sqrt{3}(430)$

Divide both sides by 2.

$\frac{dy}{dt}=\frac{\sqrt{3}(430)}{2}$

$\frac{dy}{dt}=372.390923627$

$\frac{dy}{dt}\approx 372$

Therefore the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

6 0

3 years ago

Which triangles are congruent by AAS?

netineya [11]

For two triangles to be congruent by AAS:
1- Two angles in the first triangle must be equal to two angles in the second triangle
2- A non included side in the first triangle is equal to a non included side in the second triangle

Now, let's check our options. We will find that:
For the two triangles UTV and ABC:
angle T = angle A
angle V = angle C
TU (non-included between angles T & V) = AB (non-included between angles A & C)

Therefore, we can conclude that:
Triangles ABC and UTV are congruent by AAS

8 0

2 years ago

Read 2 more answers

Somebody please help me answer this

nata0808 [166]

Answer:

Step-by-step explanation:

When somethin is decreasing you would subtract the percent it was decreasing by, by 1.

.024 - 1 = .976

7 0

2 years ago

Plzz help asap.......

SSSSS [86.1K]

Answer:

29.78 feet

Step-by-step explanation:

4 0

3 years ago