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2 years ago

Find the general indefinite integral. (use c for the constant of integration. )

Mathematics

1 answer:

Vlada [557]2 years ago

7 0

The indefinite integral will be $\int (7x^2+8x-2)dx=\dfrac{7x^3}{3}+4x^2-2x+C)$

<h3 /><h3>what is indefinite integral?</h3>

When we integrate any function without the limits then it will be an indefinite integral.

General Formulas and Concepts:

Integration Rule [Reverse Power Rule]:

$\int x^ndx=\dfrac{x^{n+1}}{n+1}}+C$

Integration Property [Multiplied Constant]:

$\int cf(x)dx=c\int f(x)dx$

Integration Property [Addition/Subtraction]:

$\int [f(x)\pmg(x)]dx=\int f(x)dx\pm \intg(x)dx$

[Integral] Rewrite [Integration Property - Addition/Subtraction]:

$\int (7x^2+8x-2)dx=\int 7x^2dx+\int 8xdx -\int 2dx$

[Integrals] Rewrite [Integration Property - Multiplied Constant]:

$\int (7x^2+8x-2)dx=7 \int x^2dx+ 8 \int xdx -2\int dx$

[Integrals] Reverse Power Rule:

$\int (7x^2+8x-2)dx= 7(\dfrac{x^3}{3})+8(\dfrac{x^2}{2})-2x+C$

Simplify:

$\int (7x^2+8x-2)dx= \dfrac{7x^3}{3}+4x^2-2x+C$

So the indefinite integral will be $\int (7x^2+8x-2)dx= \dfrac{7x^3}{3}+4x^2-2x+C$

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Shawna found coins worth $4.32. One- fourth of the found coins are pennies and one- sixth are quarters. The number of nickels fo

eduard

Answer:

8 quarters

12 nickels

12 pennies

16 dimes

Step-by-step explanation:

The trick is to ASSUME the dimes.

Start by thinking about the three explicit premises and the one implicit premise.

1) Quarters = 1/6 of total coins

2) Nickels = 1.5 the number of quarters

3) Pennies = 1/4 of total coins

Also, ASSUME an unknown number of dimes to fill in the gap.

48 Total Coins = $4.32

⅙ of 48 = 8 8 quarters = $2.00

1.5 x 8 = 12 12 nickels = $0.60

¼ of 48 = 12 12 pennies = $0.12

Add those coins = $2.72

Take the known total and minus this sum.

$4.32 - $2.72 = $1.60

16 dimes = $1.60

Thus, you have the following:

8 quarters (⅙ of total coins) $2.00

12 nickels (1.5 the number of quarters) $0.60

12 pennies (¼ the number of total coins) $0.12

+ 16 dimes $1.60

Total Coins 48 $4.32

8 0

3 years ago

Which transformations can be used to map a triangle with vertices A(2, 2), B(4, 1), C(4, 5) to A’(–2, –2), B’(–1, –4), C’(–5, –4

jek_recluse [69]

Notice that every pair of point (x, y) in the original picture, has become (-y, -x) in the transformed figure.

Let ABC be first transformed onto A"B"C" by a 90° clockwise rotation.

Notice that B(4, 1) is mapped onto B''(1, -4). So the rule mapping ABC to A"B"C" is (x, y)→(y, -x)

so we are very close to (-y, -x).

The transformation that maps (y, -x) to (-y, -x) is a reflection with respect to the y-axis. Notice that the 2. coordinate is same, but the first coordinates are opposite.

ANSWER:

"a 90 clockwise rotation about the origin and a reflection over the y-axis"

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3 years ago

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Translate the word phrase into a variable expression. five more than the product of x and 6

Natasha_Volkova [10]

Answer:

6X+5

Step-by-step explanation:

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2 years ago

A line passes through the points (-6, 4) and (-2, 2). Which is the equation of the line?

cupoosta [38]

Answer:

$\displaystyle y=-\frac{1}{2}x+1$

Step-by-step explanation:

The equation of any line in slope-intercept form is:

y=mx+b

Being m the slope and b the y-intercept.

Assume we know the line passes through points A(x1,y1) and B(x2,y2). The slope can be calculated with the equation:

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Two points are given: (-6,4) and (-2,2). Calculating the slope:

$\displaystyle m=\frac{2-4}{-2+6}=\frac{-2}{4}=-\frac{1}{2}$

The equation of the line is, so far:

$\displaystyle y=-\frac{1}{2}x+b$

To calculate the value of b, we use any of the given points, for example (-6,4):

$\displaystyle 4=-\frac{1}{2}(-6)+b$

$\displaystyle 4=3+b$

Solving:

b = 1

The equation of the line is:

$\boxed{\displaystyle y=-\frac{1}{2}x+1}$

We can see none of the choices is correct.

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3 years ago

Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$