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3 years ago

I WILL MARK BRAINLIST

1 answer:

Taya2010 [7]3 years ago

8 0

Okay so to get this answer u have too Take 17 and subtract 2 This equals 15, So the answer would be 15, This is how I get my answers at least

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Help meh plzzzzzzzzzzzz!!!!!!!!!

fgiga [73]

Answer:

1. a 2. d 3. c 4. d 5. b

Step-by-step explanation:

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3 years ago

53x – 4 = 30 O 10/3 O 40/3 40 10

Alekssandra [29.7K]

Answer:

53x - 4 = 30

53x = 30 +4

53x = 34

x= 34/53

I don't see a answer that matches the answer to the problem that you stated. Check problem.

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Which relationship describes angles 1 and 2?

In-s [12.5K]

Answer:

Supplementary Angles

Adjacent Angles

Step-by-step explanation:

Angles 1 and 2 share a vertex and form a straight line. These angles are called supplementary angles. They add to 180 degrees and form a straight line. They are adjacent as well because they share a vertex and side.

3 0

3 years ago

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.

Ben:

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

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

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

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

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 $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

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:

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