Solving for systems: {y=3x+7 & y=5x-7

2 answers:

8 0

$\left\{\begin{array}{ccc}y=3x+7\\y=5x-7\end{array}\right\\\\substitute:\\\\5x-7=3x+7\\5x-3x=7+7\\2x=14\ \ \ \ /:2\\x=7\\\\y=3\cdot7+7=21+7=28\\\\Answer:x=7\ and\ y=28.$

irina1246 [14]3 years ago

5 0

Y = 3x + 7
y = 5x-7

Now, we know that y = y

thus,

3x + 7 = 5x-7
3x - 5x = -7 -7

= -2x = -14

thus, -x = -14/-2

thus, x = 7.

Substituting for y,

y = 3x + 7
= 3 x 7 + 7

= 28.

Thus, x = 7, y =28

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Find the slope and the y-intercept of the equation 4x + 5y = -9

lutik1710 [3]

Answer:

In your equation 4x+5y=-9, you've got to but your equation in slope intercept form which is y=mx+b.

First we need to get the y by itself.

Subtract 4x from each side to get:

5y=-4x-9

Next we want to divide both sides by 5.

y=-4/5x-9/5

Your slope is m which in our case -4/5.

Then our y-intercept is b or -9/5.

Your answer would then be C. slope=-4/5 and y-intercept=-9/5.

Hope this helps ;)

7 0

3 years ago

Bonita said that the product of 5/6 times 1 2/3 is 7/3 how can you tell her answer is wrong?

kotegsom [21]

The answer is really 1 5/6 because 5/6 + 2/3=5/6 NOT 7/3 so of course she is wrong

6 0

3 years ago

Read 2 more answers

The height of a ball thrown into the air after t seconds have elapsed is h = −16t2 + 40t + 6 feet. What is the first time, t, wh

sp2606 [1]

<h3>The first time when the ball will reach a height of 20 feet is 0.42 seconds</h3>

Solution:

Given that,

The height of a ball thrown into the air after t seconds have elapsed is:

$h = -16t^2 + 40t + 6$

What is the first time, t, when the ball will reach a height of 20 feet?

Substitute h = 20

$20 = -16t^2 + 40t + 6\\\\-16t^2 + 40t + 6 -20 = 0\\\\-16t^2 + 40t -14 = 0\\\\16t^2 -40t + 14 = 0\\\\8t^2 -20t + 7=0$

Solve by quadractic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=8,\:b=-20,\:c=7$

$t = \frac{-\left(-20\right)\pm \sqrt{\left(-20\right)^2-4\cdot \:8\cdot \:7}}{2\cdot \:8}\\\\t = \frac{20 \pm \sqrt{176}}{16}\\\\t = \frac{20 \pm 4\sqrt{11}}{16}\\\\t = \frac{ 5 \pm \sqrt{11}}{4}\\\\We\ have\ two\ solutions\\\\ t=2.07915, \:t=0.42084$