What is the image of (−3,4) after the transformation ry=x?

Brad owns 3 surfboards and 2 wet suits. If he takes one surfboard and one wet suit to the beach, how many different combinations

Reptile [31]

Answer:

1

Step-by-step explanation:

He only has 1 wetsuit and 1 surfboard.

6 0

1 year ago

Alvin and Bala has 26 stickers.Alvin had 8 more stickers thab Bala.How many stickers did Bala have?

Brilliant_brown [7]

Bala had 9 stickers

You could set this up as a equation. Because there was a total of 26 you would for sure put =26. Next you are told Alvin has 8 MORE than Bali, therefore you would be adding the unknown value of Bala by 8. This could be represented as x+8=26. Now that you have x added to 8 you need to add another x to the equation to fully represent the problem since Alvin has 8 more stickers than Bala does. The new equation would become 2x+8=26.

You must now isolate x by first subtraction 8 from both sides which will leave you with 2x=18. Then you divide on both sides by 2 and will leave you with x=9

3 0

2 years ago

A rectangular table is positioned in a 10-foot by 10-foot room as shown.

makkiz [27]

Answer:

40 and 20

Step-by-step explanation:

3 0

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.

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

1 year ago

Draw a bar model that shows how the number of hours in March compares with the number of hours in February of this year.

gogolik [260]

Okay so first we need to find how many days are in March and February. March has 31 days and because this year was a leap year February has 29 days.

The next step is to convert days to hours.
March: 31x24=744
February: 29x24=696

Now its time to graph

8 0

2 years ago

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