If there's a sale and the original price is 140 and it's one fifth off how much is the sale?

Describe how you would write 5.04 X 10^6 in standard form.

Margarita [4]

You move the decimal point 6 places to the right because the exponent is positive.

5.04 x 10^6 in standard form is 5040000.

3 0

3 years ago

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There are 18 girls in mr. Brimley’s math class. If 60% of the students are girls, how many students are in mr. brimley’s math cl

m_a_m_a [10]

Answer:

30

Step-by-step explanation:

3 0

3 years ago

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What is the point slope equation of a line with slope -5 that contains the point (6,3)

ehidna [41]

Step-by-step explanation:

Given

Slope (m) = -5

Point

(x1 , y1) = ( 6 , 3)

So the equation is

y - y1 = m ( x - x1)

y - 3 = -5 ( x - 6)

y - 3 = -5x +30

5x + y = 30 + 3

5x + y = 33

5x + y - 33= 0

Which is the required equation.

Hope it will help you :)

3 0

3 years ago

Solve and graph the absolute value inequality: |2x + 4| > 14.

Aloiza [94]

Answer:

"number line with open circles on negative 9 and 5, shading going in the opposite directions."

Step-by-step explanation:

Your inequality doesn't include an equal sign so there will be no closed holes. It will only be open holes.

|u|>14 means that the number u has to be greater than 14 or less than -14. These numbers I describe just now all have a distance greater than 14 from 0.

So |u|>14 implies u>14 or u<-14.

But we are solving |2x+4|>14 so this implies we have 2x+4>14 or 2x+4<-14.

2x+4>14

Subtract 4 on both sides:

2x >10

Divide both sides by 2:

x >5

2x+4<-14

Subtract 4 on both sides:

2x <-18

Divide both sides by 2:

x <-9

So our solution is x>5 or x<-9.

Graphing!

~~~~~~~O O~~~~~~~~

-----------(-9)---------------------------------(5)---------------

So we shaded to the right of 5 because our inequality says x is bigger than 5.

We shaded to the left of -9 because our inequality says x is less than -9.

4 0

3 years ago

A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,

algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}$

Where $v_i$ is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is $[0, \pi/2]$ . The critical points of the function are those who make $d'(\theta)=0$ :

$d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}$

$d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...$

The critical value inside the interval is $\pi/4$ .

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft$

The second step is to find the values of the function at the endpoints of the interval:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft$

The biggest value of f is gived by $\pi/4$ , therefore $\pi/4$ is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

4 0

3 years ago