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3 years ago

Simplify -2[-6(-4 7)] 36 -24 -36

Mathematics

2 answers:

7nadin3 [17]3 years ago

3 0

-2(-53)×36-24-36 》3816-60=3756

Bess [88]3 years ago

3 0

36 is the answer most definitely

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Given the points P(2,-1) and Q(-9,-6), what are the coordinates of the point on directed line segment PQ that partitions PQ in t

gayaneshka [121]

ANSWER

$( - \frac{23}{5} , - 4)$

EXPLANATION

Given the points P(2,-1) and Q(-9,-6),the coordinates of the point that partition the directed line segment PQ in the ratio 3:2 is given by

$x = \frac{ mx_2+nx_1}{m + n}$

$y= \frac{ my_2+ny_1}{m + n}$

Where m=3 and n=2

$x = \frac{ 3 ( - 9)+2(2)}{3+ 2}$

$x = \frac{ - 23}{5}$

$y= \frac{ 3( - 6)+2( - 1)}{3 + 2}$

$y= \frac{ - 20}{5} = - 4$

The point is

$( - \frac{23}{5} , - 4)$

4 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.

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

<em><u>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 </u></em> $h(t)=0$ <em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

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:

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