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3 years ago

Please help u guys!!!!

Mathematics

1 answer:

damaskus [11]3 years ago

3 0

-1.....if the number on the left is smaller than the number on the right change the the number on the right to equal the same on the left then however many numbers down you went make that a negative and that’s your answer

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Is 4/5 bigger than 0.75

hjlf

4/5 is greater than 0.75. That is because, if you turn 4/5 into a decimal, them that decimal will be equal to 80/100. 0.75 is 75/100 as a fraction. 80/100 (4/5 simplified) compared to 75/100 is greater. The answer is yes.

5 0

4 years ago

Find the slope of the line that passes through (10,8) and (8,0).

N76 [4]

Answer:

The slope of the line that passes through points ( -10, -8 ) and ( -8, -16 ) is -4.

Explanation:

Points given = ( -10, -8) and ( -8, -16)

Slope = ?

( -10, -8 ) : x1 = -10 and y1 = -8

( -8, -16 ) : x2 = -8 and y2 = -16

We know,

slope = y2 - y1 / x2 - x1

Slope = -16 - ( -8) / -8 - (-10)

slope = -16 + 8 / -8 + 10

slope = -8 / 2

slope = -4

Therefore, slope of the line that passes through points ( -10, -8 ) and ( -8, -16 ) is -4.

Step-by-step explanation:

3 0

3 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:

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