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3 years ago

What is the solution to the system of equations?

Mathematics

2 answers:

DaniilM [7]3 years ago

5 0

No solution would be the answer.

german3 years ago

3 0

ANSWER

The given system of equations has no solution.

EXPLANATION

We write the two equations in the slope-intercept form,

$y = mx + c$

The first equation is
$2x - y = 7$

We make y the subject to obtain,

$y = 2x - 7 - - (1)$

The second equation is already in slope intercept form.

$y = 2x + 3 - - (2)$

We can see that the two equations have the same slope but different y-intercept. This implies that, the two equations will never intersect. Therefore the system of equations has no solution.

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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

3 years ago

In a causal-comparative study, what is the difference between independent variables and dependent variables? independent variabl

AlladinOne [14]

Answer: independent variables cause the effect measured in the dependent variables

Step-by-step explanation: In a causal - comparative study, the dependent variables refers to the variable whose changes is being measured or observed. The changes or alteration of the dependent variable is induced by the different values of the independent variable. In causal - comparative study, the different independent variables is assumed to have a direct impact on the output or values of the dependent variable which is measured by the experimenter. Therefore, the independent variable does not change, but causes the observed changes noticed in the dependent variable.

7 0

3 years ago

25 and 65 thousands as a decimal

slava [35]

The answer would be 25.65

6 0

3 years ago

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What is the slope of the line that passes through the points (-7, -7) and (-7, 1)? Write your answer in simplest form.

Ronch [10]

Answer:

undefined, stays on x so straight vertical - vertical line is undef slope

3 0

3 years ago

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Distribute and simplify the radicals below (√12+6)(-√8-√2)

Hatshy [7]

For this case we have the following expression:

$(\sqrt {12} +6) (- \sqrt {8} - \sqrt {2})$

We apply distributive property:

$(\sqrt {12} * - \sqrt {8}) + (\sqrt {12} * - \sqrt {2}) + (6 * - \sqrt {8}) + (6 * - \sqrt {2} ) =\\- \sqrt {96} - \sqrt {24} -6 \sqrt {8} -6 \sqrt {2} =$

We rewrite in equivalent form:

$- \sqrt {4 ^ 2 * 6} - \sqrt {2 ^ 2 * 6} -6 \sqrt {2 ^ 2 * 2} -6 \sqrt {2} =\\-4 \sqrt {6} -2 \sqrt {6} -6 * 2 \sqrt {2} -6 \sqrt {2} =$

$-4 \sqrt {6} -2 \sqrt {6} -12 \sqrt {2} -6 \sqrt {2} =\\-6 \sqrt {6} -18 \sqrt {2}$

Answer:

$-6 \sqrt {6} -18 \sqrt {2}$

8 0

3 years ago

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