All categories

3 years ago

Help ne with this math

Mathematics

1 answer:

stira [4]3 years ago

6 0

The answer is A,E,G,L,D,and H.

You can check this by doing this:

This is for A

16*4=84

12*7=64

84+64=148

You might be interested in

La adición de dos números es -17. Calcula el Numero Menor, si el mayor es -8

nasty-shy [4]

<h2> Podría ser -9</h2>

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

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 $t=0.6482$

3 0

3 years ago

The endpoints of PQ are P(2, 0) and Q (4, 2). Find the coordinates of midpoint PQ.

lawyer [7]

Answer:

Step-by-step explanation:

Parameterize the ellipse as (acos∙,bsin∙). Take points P:=(acosp,bsinp) and Q:=(acosq,bsinq) on the ellipse, with midpoint M:=(P+Q)/2.

If |PQ|=2k, then

a2(cosp−cosq)2+b2(sinp−sinq)2=4k2

The coordinates of M are

xy==a2(cosp+cosq)b2(sinp+sinq)

4 0

4 years ago

A puppy weighs 3 pounds. What is the puppy's weight in ounces

gtnhenbr [62]

The puppy weighs 48 ounces

6 0

3 years ago

Read 2 more answers

(-1, -5) (-2, -4) (-5,-1) (-4, -2)

notka56 [123]

Answer:

(-4, -2)

Step-by-step explanation:

A = (-1, 2)

Add (-3, -4) to that and you get ...

A' = (-1-3, 2-4) = (-4, -2) . . . . matches the last choice

The translation function has the effect of moving the point left 3 and down 4. You can count grid squares on the graph to see that A' ends up at (-4, -2).

8 0

3 years ago