What are two different ways of factoring:

The local zoo is filling two water tanks for the elephant exhibit. One water tank contains 30 gal of water and is filled at a co

kakasveta [241]

You first have to set up the two equations. Y is equal to the total number of gallons in a tank and x is equal to the number of hours the tanks are filling.

Since the two equations both equal y I can make them equal each other. I then solved for x as you see and got -4.

Once I found my x value I now needed to find my y. You only needed to chose one equation to solve for y, so I decided to do the first one. I plugged in -4 where my x value currently was and solved for y. I got 2.

If you have any questions please comment!! Hope this helps!!

3 0

3 years ago

I need help with this question !

lidiya [134]

Answer:

3x -2 ≤ 2x +1

3x ≤ 2x +3

x ≤ 3

so its {x|x∈ℝ, x ≤ 3}

7 0

2 years ago

Read 2 more answers

The number of bees that visit a plant is 600 times the number of years the plant is alive, where t represents the number of year

zavuch27 [327]

C for sure, lock it down

7 0

3 years ago

Which equation represents the data in

klasskru [66]

Answer:

y = 2x - 4

Step-by-step explanation:

The equations are put in slope intercept form

Slope intercept form: y = mx + b

Where m = slope and b = y intersect

So in order to find the equation of the data represented by the table we will have to find the slope and y intercept

Let's begin!

First let's find the slope

We can find the slope by using the slope formula

m = (y2 - y1) / (x2 - x1) where the x and y values are derived from coordinates from the table

The points chosen may vary but I have chosen the points (0,-4) and (1,-2)

Now that we have chosen the points we will use to find the slope let's define the variables

remember coordinates are written like this: (x,y)

The x value of the second coordinate is 1 so x2 = 1

The x value of the first coordinate is 0

So x1 = 0

The y value of the second coordinate is -2 so y2 = -2

The y value of the first coordinate is -4

So y1 = -4

Now that we have defined each variable let's plug in the values into the formula

Formula: m = (y2 - y1) / (x2 - x1)

Variables: x2 = 1, x1 = 0, y2 = -2, y1 = -4

Substitute values

m = (-2 - (-4) / ( 1 - 0 )

Evaluate

The negative signs cancel out on top and it changes to +4

m = (-2 + 4)/(1-0)

Add top values

m = 2/(1-0)

Subtract bottom numbers

m = 2/1

Simplify fraction

m = 2

So we can conclude that the slope (m) = 2

Now let's find the y intercept or "b"

The y intercept is the value of y when x = 0

If you look at the table when x = 0 y = -4 meaning that the y intercept or "b" is -4

Now that we have found everything let's find the equation of the data represented by the table

The equation is in slope intercept form

y = mx + b

Define variables

m = 2 and b = -4

Substitute values

y = 2x - 4

The equation is y = 2x - 4

4 0

2 years ago

A. Evaluate ∫20 tan 2x sec^2 2x dx using the substitution u = tan 2x.

irakobra [83]

Answer:

The integral is equal to $5\sec^2(2x)+C$ for an arbitrary constant C.

Step-by-step explanation:

a) If $u=\tan(2x)$ then $du=2\sec^2(2x)dx$ so the integral becomes $\int 20\tan(2x)\sec^2(2x)dx=\int 10\tan(2x) (2\sec^2(2x))dx=\int 10udu=\frac{u^2}{2}+C=10(\int udu)=10(\frac{u^2}{2}+C)=5\tan^2(2x)+C$ . (the constant of integration is actually 5C, but this doesn't affect the result when taking derivatives, so we still denote it by C)

b) In this case $u=\sec(2x)$ hence $du=2\tan(2x)\sec(2x)dx$ . We rewrite the integral as $\int 20\tan(2x)\sec^2(2x)dx=\int 10\sec(2x) (2\tan(2x)\sec(2x))dx=\int 10udu=5\frac{u^2}{2}+C=5\sec^2(2x)+C$ .

c) We use the trigonometric identity $\tan(2x)^2+1=\sec(2x)^2$ is part b). The value of the integral is $5\sec^2(2x)+C=5(\tan^2(2x)+1)+C=5\tan^2(2x)+5+C=5\tan^2(2x)+C$ . which coincides with part a)

Note that we just replaced 5+C by C. This is because we are asked for an indefinite integral. Each value of C defines a unique antiderivative, but we are not interested in specific values of C as this integral is the family of all antiderivatives. Part a) and b) don't coincide for specific values of C (they would if we were working with a definite integral), but they do represent the same family of functions.

3 0

3 years ago