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3 years ago

Gary is using an indirect method to prove that segment DE is not parallel to segment BC in the triangle ABC shown below:

Mathematics

1 answer:

ahrayia [7]3 years ago

7 0

Answer:

That Ratios aren't equal.

Step-by-step explanation:

Given: DE is || to BC.

So in order to make this false then we have to say that the sides aren't proprtional, making it not possible to get the ratios equal.

(See the Triangle proprtionality theorem or the triangle midsegment theorem)

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a math student is standing 40 feet from the base of a water tower. the angle of elevation from her horizontal line of sight is 7

Lerok [7]

9514 1404 393

Answer:

128 ft

Step-by-step explanation:

The height of the tower above eye-height is related to the angle by ...

Tan = Opposite/Adjacent

tan(72°) = (tower height above eye height)/(40 ft)

tower height above eye height = (40 ft)×tan(72°) = 123 ft

Since eye height is 5 ft, the total height of the tower from the ground is ...

tower height = eye height + height above eye height

tower height = 5 ft + 123 ft

tower height = 128 ft

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2 years ago

Solve for x. 1n(5x-3)=2

fredd [130]

So sorry about my handwriting

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3 years ago

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mariarad [96]

Answer: A

Step-by-step explanation:

Each gallon costs 3 dollars.

8(gallons) times $3(for each gallon) = 24(dollars).

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3 years ago

The amount of money it takes to fill up a gas tank varies directly with the number of gallons bought. It cost $21.87 to fill up

Kobotan [32]

I believe the answer is C. 2.43

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3 years ago

In physics, if a moving object has a starting position at so, an initial velocity of vo, and a constant acceleration a, the

emmasim [6.3K]

Answer:

$a=\frac{2S -2v_ot-2s_o}{t^2}$

Step-by-step explanation:

We have the equation of the position of the object

$S = \frac{1}{2}at ^2 + v_ot+s_o$

We need to solve the equation for the variable a

$S = \frac{1}{2}at ^2 + v_ot+s_o$

Subtract $s_0$ and $v_0t$ on both sides of the equality

$S -v_ot-s_o = \frac{1}{2}at ^2 + v_ot+s_o - v_ot- s_o$

$S -v_ot-s_o = \frac{1}{2}at ^2$

multiply by 2 on both sides of equality

$2S -2v_ot-2s_o = 2*\frac{1}{2}at ^2$

$2S -2v_ot-2s_o =at ^2$

Divide between $t ^ 2$ on both sides of the equation

$\frac{2S -2v_ot-2s_o}{t^2} =a\frac{t^2}{t^2}$

Finally

$a=\frac{2S -2v_ot-2s_o}{t^2}$

5 0

3 years ago