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natta225 [31]
2 years ago
6

Right angle FCD intersects Line A B and Ray C E at point C. AngleFCE is congruent to AngleECD. AngleECD is complementary to Angl

eDCB. A horizontal line has points A, C, B. Right angle F C D intersects the line at point C. A line extends from point C vertically to point E. Angle A C F is 45 degrees and angle F C D is 90 degrees. Which statement is true about AngleDCB and AngleACF? They are congruent and complementary. They are congruent and supplementary. They are complementary but not necessarily congruent. They are supplementary but not necessarily congruent.
Mathematics
2 answers:
AlexFokin [52]2 years ago
5 0

Answer:

A: They are congruent and complementary.

Step-by-step explanation:

Just took the test hope this helps :D

Vladimir [108]2 years ago
3 0

Answer:

A

Step-by-step explanation:

it was correct on edge2020 so

(shrug)

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Ivanshal [37]
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2 years ago
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Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be
Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of t=0.6482

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

h(t)=-16t^2 +v_{o}t+s       Eqn(1).

Where:

h(t) : is the height range as a function of time

v_{o}   : is the initial velocity

s     : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

h(t)=-16t^2 + v_{o}t + 40        Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

<u>Ben:</u>

We know that v_{o}=60ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+60t+40        Eqn.(3)

<u>Josh:</u>

We know that v_{o}=48ft/s

Thus Eqn. (2) becomes:

h(t)=-16t^2+48t+40       Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em>h(t)=0<em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

Thus we have for Ben, Eqn. (3) gives:

h(t)=0-16t^2+60t+40=0

Using the quadratic expression to find the roots of the quadratic we have:

t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec

Since time can only be positive we reject the t_{2} solution and we keep that Ben's book took t=4.3276 seconds to reach the ground.

Similarly solving for Josh we obtain

t_{1}=3.6794sec\\t_{2}=-0.6794sec

Thus again we reject the negative and keep the positive solution, so Josh's book took t=3.6794 seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

  • Josh's textbook reached the ground first
  • Josh's textbook reached the ground first by a difference of t=0.6482

3 0
2 years ago
The volumes of two similar figures are 343 mm3 and 512 mm3. If the surface area of the larger figure is 192 mm2, what is the sur
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In geometry, similar figures are those whose ratios of the  corresponding sides are equal and the corresponding  angles are congruent. In relation to the volume, we determine first the cube roots of the given and find the ratio as shown below.
 
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The square of this ratio is the ratio of the areas of the figure. If we let x be the area of the smaller figure then, 
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The value of x from the equation is 147 mm². 

The area therefore of the smaller figure is 147 mm².
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Step-by-step explanation:

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4(2 - x) > -2x - 3(4x + 1)
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Therefore, x = 0 and x = 10 zre solutions to the inequality.
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2 years ago
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