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3 years ago

What theorem can be used to prove that the two triangles are congruent?

Mathematics

1 answer:

choli [55]3 years ago

6 0

The answer is D)
SSS

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If it's 6:15 PM what time is it gonna be in 45 minutes

beks73 [17]

Answer:

7 pm is the answer

15±45=60

then 7pm

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3 years ago

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Consider the diagram.

Dima020 [189]

Answer:

AB = 9

Step-by-step explanation:

Given that line, l is a perpendicular bisector of line segment AC. It intersects line segment AC at point B.

So, we can conclude that AB = BC ............ (1)

Also given that the length of line segment BC is 9.

Now, from equation (1) we can say that, AB = 9.

Therefore, the length of segment AB is 9. (Answer)

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3 years ago

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Write the equation of the parabola in vertex form.vertex (0,2), point (1,-5)

charle [14.2K]

Y=a(x-h)^2+k
-5=a(1-0)^2+2
-5=a+2
-3=a

y=-3x^2+2

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3 years ago

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Rectangular field that measure 2 feet by 18 inches

UkoKoshka [18]

Answer:

432 square inches

Step-by-step explanation:

thats the area, perimiter is 84 inches

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3 years ago

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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

4 years ago