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3 years ago

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the area of a rectangle is given by the equation y=x^2-9x+14 and the length of the rectangle is x-2. which of the following expr

essions represents the width of the rectangle?

1 answer:

Rudik [331]3 years ago

3 0

Divide x2 - 9x +14 by x-2
You get x - 7

Check your work by multiplying x-2 and x-7
Since you get x2 - 9x +14,

x-7 is the answer.

You might be interested in

What is 5-6x=2+7x step by step?

HACTEHA [7]

The first thing you do is subtract 2 from both sides to move the 2 to the left side
3-6x=7x
now add 6x to both sides to move -6x to the right
3=13x
divide by 13 to eliminate the 13 on the right side
3/13

4 0

3 years ago

Read 2 more answers

Can anyone please help me in this

Alik [6]

These problems are called systems of equations. Basically you have two linear equations and you need to find the values for x and y. In other words, all these equation are lines and our answer will be the exact point that the pair of lines intersect. For example, if we get x=1 and y=2 the lines will intersect at point (1,2). Now that you have some background knowledge here comes the tricks and tactics kid.

We know that we can solve one variable equation easily. For example...

x+1=2

x=1 obviously

Cause we have two variables x and y it is not possible to find a solution. For example, in the equation x+y=10, x=1 when y=9 and x=2 when y=8. There is not correct answer.

So what can we do? We have to make a two variable equation into a one variable equation.

There are two ways to do this: substitution and elimination. I will create a sample problem and then solve it using both methods.

x+y=2

2y-y=1

3)

-3x-5y=-7 -----> -12x-20y=-28

-4x-3y=-2 ------> -12x-9y=-6

-12x-20y=-28

-(-12x-9y=-6)

---------------------

-11y=-22

y=2

-3x-5(2)=-7

-3x=3

x=-1

4) 8x+4y=12 ---> 24x+12y=36

7x+3y=10 ---> 28x+12y=40

28x+12y=40

-(24x+12y=36)

---------------------

4x=4

x=1

8(1)+4y=12

4y=4

y=1

5) 4x+3y=-7

-2x-5y=7 ----> -4x-10y=14

4x+3y=-7

+(-4x-10y=14)

-------------------

-7y=7

y=-1

4x+3(-1)=-7

4x=-4

x=-1

6) 8x-3y=-9 ---> 32x-12y=-36

5x+4y=12 ---> 15x+12y=36

32x-12y=-36

+(15x+12y=36)

--------------------

47x=0

x=0

8(0)-3y=-9

-3y=-9

y=3

7)-3x+5y=-2

2x-2y=1 ---> x-y=1/2 ----> x=y+1/2

-3(y+1/2)+5y=-2

-3y-1.5+5y=-2

2y=-0.5

y=0.25

2x-2(0.25)=1

2x=1.5

x=0.75

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%281-x%5E%7B2%7D%20%29%5E%7B3%2F2%7D%20%7D%20%5C%2C%20dx" id="TexFo

Ludmilka [50]

Answer: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}$ General Formulas and Concepts:

<u>Pre-Calculus</u>

Trigonometric Identities

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Integration

Integrals
Definite/Indefinite Integrals
Integration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

U-Substitution

Trigonometric Substitution

Reduction Formula: $\displaystyle \int {cos^n(x)} \, dx = \frac{n - 1}{n}\int {cos^{n - 2}(x)} \, dx + \frac{cos^{n - 1}(x)sin(x)}{n}$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx$

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution (trigonometric substitution).</em>

Set <em>u</em>: $\displaystyle x = sin(u)$
[<em>u</em>] Differentiate [Trigonometric Differentiation]: $\displaystyle dx = cos(u) \ du$
Rewrite <em>u</em>: $\displaystyle u = arcsin(x)$

<u>Step 3: Integrate Pt. 2</u>

[Integral] Trigonometric Substitution: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[1 - sin^2(u)]^\Big{\frac{3}{2}} \, du$
[Integrand] Rewrite: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[cos^2(u)]^\Big{\frac{3}{2}} \, du$
[Integrand] Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos^4(u)} \, du$
[Integral] Reduction Formula: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{4 - 1}{4}\int \limits^a_b {cos^{4 - 2}(x)} \, dx + \frac{cos^{4 - 1}(u)sin(u)}{4} \bigg| \limits^a_b$
[Integral] Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4}\int\limits^a_b {cos^2(u)} \, du$
[Integral] Reduction Formula: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg|\limits^a_b + \frac{3}{4} \bigg[ \frac{2 - 1}{2}\int\limits^a_b {cos^{2 - 2}(u)} \, du + \frac{cos^{2 - 1}(u)sin(u)}{2} \bigg| \limits^a_b \bigg]$
[Integral] Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}\int\limits^a_b {} \, du + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg]$
[Integral] Reverse Power Rule: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}(u) \bigg| \limits^a_b + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg]$
Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3cos(u)sin(u)}{8} \bigg| \limits^a_b + \frac{3}{8}(u) \bigg| \limits^a_b$
Back-Substitute: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(arcsin(x))sin(arcsin(x))}{4} \bigg| \limits^a_b + \frac{3cos(arcsin(x))sin(arcsin(x))}{8} \bigg| \limits^a_b + \frac{3}{8}(arcsin(x)) \bigg| \limits^a_b$
Simplify: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x)}{8} \bigg| \limits^a_b + \frac{x(1 - x^2)^\Big{\frac{3}{2}}}{4} \bigg| \limits^a_b + \frac{3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b$
Rewrite: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x) + 2x(1 - x^2)^\Big{\frac{3}{2}} + 3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Book: College Calculus 10e

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3 years ago

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Which equation is equivalent to ax+b=c

alexgriva [62]

Answer: b = c-ax

hope this helps:)

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3 years ago

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3x+y=9 and 3x=9-y solve each system by elimination to prove your answer, I am stuck on this one

Svetllana [295]

Infinitely many solutions.I will try to solve your system of equations.<span><span><span><span>3x</span>+y</span>=9</span>;<span><span>3x</span>=<span><span>−y</span>+9</span></span></span>Step: Solve<span><span><span>3x</span>+y</span>=9</span>for y:<span><span><span><span>3x</span>+y</span>+<span>−<span>3x</span></span></span>=<span>9+<span>−<span>3x</span></span></span></span>(Add -3x to both sides)<span>y=<span><span>−<span>3x</span></span>+9</span></span>Step: Substitute<span><span>−<span>3x</span></span>+9</span>foryin<span><span><span>3x</span>=<span><span>−y</span>+9</span></span>:</span><span><span>3x</span>=<span><span>−y</span>+9</span></span><span><span>3x</span>=<span><span>−<span>(<span><span>−<span>3x</span></span>+9</span>)</span></span>+9</span></span><span><span>3x</span>=<span>3x</span></span>(Simplify both sides of the equation)<span><span><span>3x</span>+<span>−<span>3x</span></span></span>=<span><span>3x</span>+<span>−<span>3x</span></span></span></span><span>(Add -3x to both sides)

so ur answer is
</span><span>Infinitely many solutions.</span><span>

</span>

7 0

3 years ago