Please help i don’t know how to do this

6 pints of a 20 percent solution of alcohol in water are mixed with 4 pints of a 10 percent alcohol in water solution. The perce

andreev551 [17]

Total Volume would be 6 + 4 = 10 pints
Let's calculate the volume of alcohol in final solution:
20% * 6 = 1.2 pints
10% * 4 = .4 pints
Total = 1.6 pints of alcohol in 10
Percentage = (1.6/10) * 100 = 16 per cent.
answer is A

4 0

1 year ago

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Find x rounded to one decimal place.

8090 [49]

There is little information please state more points.

7 0

1 year ago

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The function f describes the value of a sculpture after t years. The function g describes the value of a painting by the same ar

Anna11 [10]

H(t) = f(t) - g(t)
= 500(1.2)^t - 380(1.15)^t
Taking out the greatest common factor, which is 20:
= 20[(25)(1.2)^t - (19)(1.15)^t]
This is the third choice.

Note that you cannot subtract the bases of the exponents, for example (1.2^t - 1.15^t) cannot be simplified into something like 0.05^t.

6 0

2 years ago

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Quadrilateral CAMP below is a rhombus. the length PQ is (x+2) units, and the length of QA is (3x-14) units. Which statements bes

valentinak56 [21]

Answer:

<h3>Q cuts the diagonal PA into 2 equal halves, since the diagonals of rhombus meet at right angles.</h3><h3>The value of x is 8.</h3>

Step-by-step explanation:

Given that Quadrilateral CAMP below is a rhombus. the length PQ is (x+2) units, and the length of QA is (3x-14) units

From the given Q is the middle point, which cut the diagonal PA into 2 equal halves.

By the definition of rhombus, diagonals meet at right angles.

Implies that PQ = QA

x+2 = 3x - 14

x-3x=-14-2

-2x=-16

2x = 16

dividing by 2 on both sides, we will get,

$x =\frac{16}{2}$

<h3>∴ x=8</h3><h3>Since Q cuts the diagonal PA into 2 equal halves, since the diagonals of rhombus meet at right angles we can equate x+2 = 3x-14 to find the value of x.</h3>

The line segment $\overrightarrow{PA}=\overrightarrow{PQ}+\overrightarrow{QA}$

$\overrightarrow{PA}=x+2+3x-14$

$=4x-12$

$=4(8)-12$ ( since x=8)

$=32-12$

$=20$

<h3>∴ $\overrightarrow{PA}=20$ units</h3>

3 0

1 year 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:

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