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2 years ago

Please help me, please

Mathematics

2 answers:

LiRa [457]2 years ago

8 0

Answer: D: -2
explanation:
1. divide fractions inside brackets by multiplying by the reciprocal
1/4 • 4/7
= 4/28
2. multiply 4/28 by 7/5
= 28/140 which can be reduced to 1/5
3. multiply by -10
= 1/5 • -10
= -2

Svetlanka [38]2 years ago

5 0

Answer:

-2

Step-by-step explanation:

7/5(1/4 divide 7/4) (-10)

Multiply the reciprocal of that fraction

7/5x(1/4x4/7)x(-10)

four and four cancels out

7/5x1/7x(-10)

multiplying a positive and a negative equals a negative

-7/5x1/7x10

seven cancels out and u get

-1/5x10

divide 5 both sides and u get

-2

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The percent of concentration of a certain drug in the bloodstream x hours after the drug is administered is given by K(x) 5x/x^2

Olegator [25]

Given :

The percent of concentration of a certain drug in the bloodstream x hours after the drug is administered is given by $K(x) = \dfrac{5x}{x^2+9}$ .

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Solution :

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2 years ago

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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:

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