5а - 2ь – 3c = 35 please answer

If each one of the 15 students raises the same amount, let's define this amount by the variable X, then the total amount they will collect is 15 times X, or:

15*X

And the minimum amount they need to collect is $800

Then we will have the inequality:

15*X ≥ $800

Now we can divide both sides by 15 to get the minimum amount that each student needs to raise:

(15*X)/15 ≥ $800/15

X ≥ $54.33...

Where we should round up the right number to $54.35, because we can not have a $0.0333... and if we round down, they would collect less than $800.

Then the inequality is:

X ≥ $54.35

This means that the minimum amount that each one of the 15 students needs to collect is $54.35

7 0

3 years ago

Match the parabolas represented by the equations with their foci.

Elenna [48]

Function 1 $f(x)=- x^{2} +4x+8$

First step: Finding when $f(x)$ is minimum/maximum
The function has a negative value $x^{2}$ hence the $f(x)$ has a maximum value which happens when $x=- \frac{b}{2a}=- \frac{4}{(2)(1)}=2$ . The foci of this parabola lies on $x=2$ .

Second step: Find the value of y-coordinate by substituting $x=2$ into $f(x)$ which give $y=- (2)^{2} +4(2)+8=12$

Third step: Find the distance of the foci from the y-coordinate
$y=- x^{2} +4x+8$ - Multiply all term by -1 to get a positive $x^{2}$
$-y= x^{2} -4x-8$ - then manipulate the constant of y to get a multiply of 4
$4(- \frac{1}{4})y= x^{2} -4x-8$
So the distance of focus is 0.25 to the south of y-coordinates of the maximum, which is $12- \frac{1}{4}=11.75$

Hence the coordinate of the foci is (2, 11.75)

Function 2: $f(x)= 2x^{2}+16x+18$

The function has a positive $x^{2}$ so it has a minimum

First step - $x=- \frac{b}{2a}=- \frac{16}{(2)(2)}=-4$
Second step - $y=2(-4)^{2}+16(-4)+18=-14$
Third step - Manipulating $f(x)$ to leave $x^{2}$ with constant of 1
$y=2 x^{2} +16x+18$ - Divide all terms by 2
$\frac{1}{2}y= x^{2} +8x+9$ - Manipulate the constant of y to get a multiply of 4
$4( \frac{1}{8}y= x^{2} +8x+9$

So the distance of focus from y-coordinate is $\frac{1}{8}$ to the north of $y=-14$
Hence the coordinate of foci is (-4, -14+0.125) = (-4, -13.875)

Function 3: $f(x)=-2 x^{2} +5x+14$

First step: the function's maximum value happens when $x=- \frac{b}{2a}=- \frac{5}{(-2)(2)}= \frac{5}{4}=1.25$
Second step: $y=-2(1.25)^{2}+5(1.25)+14=17.125$
Third step: Manipulating $f(x)$
$y=-2 x^{2} +5x+14$ - Divide all terms by -2
$-2y= x^{2} -2.5x-7$ - Manipulate coefficient of y to get a multiply of 4
$4(- \frac{1}{8})y= x^{2} -2.5x-7$
So the distance of the foci from the y-coordinate is - $\frac{1}{8}$ south to y-coordinate

Hence the coordinate of foci is (1.25, 17)

Function 4: following the steps above, the maximum value is when $x=8.5$ and $y=79.25$ . The distance from y-coordinate is 0.25 to the south of y-coordinate, hence the coordinate of foci is (8.5, 79.25-0.25)=(8.5,79)

Function 5: the minimum value of the function is when $x=-2.75$ and $y=-10.125$ . Manipulating coefficient of y, the distance of foci from y-coordinate is $\frac{1}{8}$ to the north. Hence the coordinate of the foci is (-2.75, -10.125+0.125)=(-2.75, -10)

Function 6: The maximum value happens when $x=1.5$ and $y=9.5$ . The distance of the foci from the y-coordinate is $\frac{1}{8}$ to the south. Hence the coordinate of foci is (1.5, 9.5-0.125)=(1.5, 9.375)

8 0

3 years ago

From a full 50-liter container of a 40% concentration of acid, x liters are removed and replaced with 100% acid. (A) Write the a

padilas [110]

Answer:

See explanation

Step-by-step explanation:

Given:

50 liter container

40% concentration of acid

x liters are removed and replaced with 100% acid

(A)

Acid in 50-liter container = 40% of 50 liters

= (40/100) * 50

= 0.4 * 50

= 20

So there are 20 liters amount of acid in 50 liter container.

Now x liters are removed and replaced with 100% acid

The acid in x-liter

40% of x liters = (40/100) * x = 0.4 * x

When x liters are removed then amount of acid that remained in container:

20 - 40% of x liters = 20 - (40/100) * x = 20 - 0.4 * x

This can also be written as:

(50-x) (40/100) = (40*50/100) - (40*x/100) = 2000/100 - 40x/100 = 20 - 2x/5

Since x liters replaced with 100% acid so this becomes:

20 - 40% of x liters + 100% of x liters

= 20 - (40/100) * x + 100/100 * x

= 20 - (40/100) x + (100/100)x

= 20 + (100-40/100)x

= 20 + (60/100)x

= 20 + 0.6 x

This can also be written as:

20 - 2x/5 + x = 20 + 3x/5 = 20 + 0.6 x

Hence the amount of acid in the final mixture as a function of x:

f(x) = 20 + 0.6 x

(B)

The value of x can not be greater than 50 liters and can not be lesser than 0 liters.

so, range is [0, 50] where both 0 and 50 are inclusive.

(C)

Since final mixture is 50% acid So

Acid = 50% of 50 litres =

= (50/100) * 50

= 1/2 * 50

= 50/2

= 25

Put 25 in the computed function of x

20 + 0.6 x = 25

0.6x = 25 - 20

0.6x = 5

x = 5/0.6

x = 8.333333

x = 8.3 liters

3 0

4 years ago