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3 years ago

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the

Mathematics

1 answer:

IrinaK [193]3 years ago

4 0

Answer:

1,041.9feet

Step-by-step explanation:

Given the height of the rocket expressed as

y = -16x² + 245x + 104

At maximum height, dy/dx = 0

dy/dx = -32x+245

0 = -32x+245

32x = 245

x = 245/32

x = 7.65625

Get the maximum height

Recall that;

y = -16x² + 245x + 104

Substitute the value of x;

y = -16(7.65625)² + 245(7.65625) + 104

y = -937.890625 + 1,875.78125 + 104

y = 1,041.890625feet

Hcne the maximum height to the nearest foot is 1,041.9feet

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30 treadmills to 36 elliptical machines ratio simplest form

Ivanshal [37]

Answer:

5 to 6 or 5:6 or 5/6

Step-by-step explanation:

30/36 = 15/18 = 5/6

8 0

3 years ago

Read 2 more answers

There are 125 members in the school

Igoryamba

Answer:

125x + 5,000 > 25,000

5 0

3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago

What is the slope intercept from of the equation of the line that that passes through the points (-3, 2) and (1, 5)

viva [34]

The slope-intercept form of a line:

$y=mx+b\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\to slope\\\\b\to y-intercept$

We have:

$(-3,\ 2)\to x_1=-3,\ y_1=2\\(1,\ 5)\to x_2=1,\ y_2=5$

Substitute:

$m=\dfrac{5-2}{1-(-3)}=\dfrac{3}{4}$

$y=\dfrac{3}{4}x+b$

Put the values of coordinates of the point (1, 5) to the equation of a line:

$5=\dfrac{3}{4}(1)+b\\\\5=\dfrac{3}{4}+b\qquad|-\dfrac{3}{4}\\\\b=4\dfrac{1}{4}$

Answer: $y=\dfrac{3}{4}x+4\dfrac{1}{4}$

5 0

3 years ago

A computer training institute has 625 students that are paying a course fee of $400. Their research shows that for every $20 red

Anastasy [175]

1.
If no changes are made, the school has a revenue of :

625*400$/student=250,000$

2.
Assume that the school decides to reduce n*20$.

This means that there will be an increase of 50n students.

Thus there are 625 + 50n students, each paying 400-20n dollars.

The revenue is:

(625 + 50n)*(400-20n)=12.5(50+n)*20(20-n)=250(n+50)(20-n)

3.

check the options that we have,

a fee of $380 means that n=1, thus

250(n+50)(20-n)=250(1+50)(20-1)=242,250 ($)

a fee of $320 means that n=4, thus

250(n+50)(20-n)=250(4+50)(20-4)=216,000 ($)

the other options cannot be considered since neither 400-275, nor 400-325 are multiples of 20.

Conclusion, neither of the possible choices should be applied, since they will reduce the revenue.

6 0

4 years ago