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2 years ago

What theorem shows that ΔACE ≅ ΔBCD?

Mathematics

1 answer:

Setler79 [48]2 years ago

3 0

Triangle Congruence Theorem
Hope this helps and good luckkk :)

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I just started 8th grade and I want to know if my answer is right.

mario62 [17]

Yes you do! In fact not just I. You move al l three down right after or before you move the units to the right.

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3 years ago

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What is 3 square root of 24- 4 square root of 54

erastovalidia [21]

$\sqrt[3]{24} = 2 \sqrt[3]{3} \\ \\ \sqrt[4]{64} = 2 \sqrt[4]{4} \ \ maybe \ is\ it \ this \ number ? \ 54 \ is\ \ strange$

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2 years ago

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An object is launched at 29.4 meters per

Lerok [7]

Answer:

The reasonable domain for the scenario is option 'a';

a) [0, 7]

Step-by-step explanation:

For the projectile motion of the object, we are given;

The speed at which the object is launched, v = 29.4 meters per second

The height of the platform from which the object is launched, h = 34.3 meter

The equation for the height of the object as a function of time 'x' is given as follows;

f(x) = -4.9·x² + 29.4·x + 34.3

The domain for the scenario, is given by the possible values of 'x' for the function, which is found as follows;

At the height from which the object is launched, x = 0, and f(x) = 34.3

At the ground level to which the object can drop, f(x) = 0

∴ f(x) = -4.9·x² + 29.4·x + 34.3 = 0

-4.9·x² + 29.4·x + 34.3 = 0

By the quadratic formula, we have;

x = (-29.4 ± √(29.4² - 4 × (-4.9) × 34.3))/(2 × (-4.9)

∴ x = -1, or 7

Given that time is a natural number, we have the reasonable domain for the scenario as the start time when the object is launched, t = 0 to the time the object reaches the ground, t = 7

Therefore, the reasonable domain for the scenario is; 0 ≤ x ≤ 7 or [0, 7].

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2 years ago

A street light is at the top of a 25 ft pole. A 4 ft tall girl walks along a straight path away from the pole with a speed of 6

Burka [1]

Answer:

Tip of the shadow of the girl is moving with a rate of 7.14 feet per sec.

Step-by-step explanation:

Given : In the figure attached, Length of girl EC = 4 ft

Length of street light AB = 25 ft

Girl is moving away from the light with a speed = 6 ft per sec.

To Find : Rate ( $\frac{dw}{dt}$ ) of the tip (D) of the girl's shadow (BD) moving away from th

light.

Solution : Let the distance of the girl from the street light is = x feet

Length of the shadow CD is = y feet

Therefore, $\frac{dx}{dt}=6$ feet per sec. [Given]

In the figure attached, ΔAFE and ΔADE are similar.

By the property of similar triangles,

$\frac{x}{21}=\frac{x+y}{25}$

25x = 21(x + y)

25x = 21x + 21y

25x - 21x = 21y

4x = 21y

y = $\frac{4x}{21}$

Now we take the derivative on both the sides,

$\frac{dy}{dt}=\frac{4}{21}\times \frac{dx}{dt}$

= $\frac{4}{21}\times 6$

= $\frac{8}{7}$

≈ 1.14 ft per sec.

Since w = x + y

Therefore, $\frac{dw}{dt}= \frac{dx}{dt}+\frac{dy}{dt}$

$\frac{dw}{dt}=6+1.14$

= 7.14 ft per sec.

Therefore, tip of the shadow of the girl is moving with a rate of 7.14 feet per sec.

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3 years ago

Find the slope and y intercept 2x+3y=1,470

oksano4ka [1.4K]

First we need to put it in slope intercept form (solve for y). Then we get tex]y= -\frac{2}{3} x+490[/tex]. So the slope is-2/3 and the y intercept is 490.

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2 years ago

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