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3 years ago

Order set of numbers from least to greatest 3/4, 0.72%, 69%, .63

Mathematics

2 answers:

nika2105 [10]3 years ago

6 0

Answer:

Step-by-step explanation:

0.72%, .63= 63%, 69%, 3/4

lisov135 [29]3 years ago

4 0

.63 , 69% , 0.72% , 3/4

.63 is the least

.69 is second least

.72 is the third least

3/4 (.75) is the greatest

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Solve this If you can

ASHA 777 [7]

Answer:

I cannot solve this okay

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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:

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

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:

Whose textbook reaches the ground first and by how many seconds?

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

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3 years ago

State if the two triangles are congruent. if they are state how you know.

nlexa [21]

1) The triangles are congruent by SSS.

The two tick marks indicate two pairs of congruent sides; it is evident that the third side is congruent by the way the diagram is drawn - the bases of the triangles are together and appear to be the same length.

2) The triangles are congruent by SAS.

The two pairs of tick marks indicate congruent sides, and their included angles are congruent because they are vertical angles, and vertical angles are always congruent.

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3 years ago

Which of the following relations are functions?

Alisiya [41]

{(c,e),(c,d),(c,b)} is NOT a function since the input c has multiple outputs (e,d,b). So choice B is out

{(b,b),(c,d),(d,c),(c,a)} is NOT a function either. The input 'c' corresponds to the output 'd' and 'a' at the same time. So choice C is out too

Choices A and D are the answer. They are functions since any given input corresponds to exactly one output.

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3 years ago

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Which of the following inequalities matches the graph? (1 point)

pickupchik [31]

Answer:

there were no inequalities to choose from but I believe the answer is y ≤ 4x

Step-by-step explanation:

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3 years ago