All categories

3 years ago

Kiana wants to write equations in the form y = mx + b for the lines passing through

Mathematics

1 answer:

horrorfan [7]3 years ago

3 0

Answer:

'Substitute the slope and the coordinates of point P(-3,2) in y = mx + b and then solve for b in each equation'.

Step-by-step explanation:

Kiana wants to write equations in the form y = mx + b for the lines passing through point P that are parallel and perpendicular to line q.

This equation is in the slope-intercept form where m is the slope and b is the y-intercept.

First, she finds the slope of line q and the slope of line s to be – 2. Point P is located at (-3,2).

Therefore, the step that can be used to find the y-intercept is 'substitute the slope and the coordinates of point P(-3,2) in y = mx + b and then solve for b in each equation'. (Answer)

You might be interested in

Factorise x squared - 12x + 36

slava [35]

Answer:

(x-6)(x-6) or (x-6)^2

Step-by-step explanation:

x^2-12x+36=(x-6)(x-6)

Just find two numbers that when they multiply, you get the third term,

which is 36 in this case. And when they add up, you get the second term,

which is -12 in this case. You can also check the answer by multiplying.

(x-6)(x-6)=x^2-6x-6x+36=x^2-12x+36

3 0

3 years ago

Long division method

Anton [14]

Answer: you didnt add a question or possible answer choices so nobody can answer this question

Step-by-step explanation:

Divide the ones column dividend by the divisor.

Multiply the divisor by the quotient in the right place column.

Subtract the product from the ones column.

8 0

2 years ago

If you answer this you will get 69 points but please answer asap

spin [16.1K]

Answer:

I think the answer for the first one is 3 I am not sure about the second one

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:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$