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3 years ago

What is the result of subtracting the second equation from the first? 5x-y=6 and -2x+y=8

Mathematics

1 answer:

Travka [436]3 years ago

4 0

Answer:

Solve for the first variable in one of the equations, then substitute the result into the other equation.

Point Form:

(2,−4)

Equation Form: x= 2, y = −4

Step-by-step explanation:

I did the math

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6 apples 4 bananas 3 carrots cost 2.93$ 4 apples 4 bananas and 2 carrots cost 2.22$ 8 carrots cost 0.88$ how much is each item

GarryVolchara [31]

Answer:

Step-by-step explanation:

carrot = 0.11 $ .ie 0.88/8

3 0

2 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 $t=0.6482$

3 0

3 years ago

The points P and Q in the following figure define the line segment (PQ⎯⎯⎯⎯⎯) P Q . The steps involved in using a compass to cons

kenny6666 [7]

You haven't provided the steps.

mathisfun.com/geometry/construct-linebisect.html

Here is a useful link to the correct steps. The instructions may not be exactly the same but I think you can do it.

3 0

3 years ago

The table shows the number of cups of water required when cooking different amounts of rice.

Harman [31]

Answer-

The amount of water is the dependent value.

Solution-

Plotting the graph from the given data points, it comes out to be a straight line with positive slope. So, both the variables i.e amount of rice and amount of water are in a linear relationship. They are directly proportional to each other.

As the amount of rice increases, the water needed to cook the rice also increases. Plotting a graph taking amount of rice in abscissa and amount of water in ordinate, a straight line was obtained.

Here, amount of rice is the independent variable and amount of water is dependent variable as it depends on the amount of rice.