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3 years ago

6(5k-8)-20=11(2k-3)+3k

Mathematics

2 answers:

DochEvi [55]3 years ago

6 0

The answer for this question is k=7

stira [4]3 years ago

4 0

30k-68=25k-33
30k-25k=-33+68
5k=35
k=7

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4(5x – 9) = -2(x + 7) solve and justify each step

Alexandra [31]

Step-by-step explanation:

4(5x-9)=-2(x+7)

4*5x-4*9=-2*x-2*7

20x-36=-2x-14

20x+2x=-14+36

22x=22

x=22/22

x=1

5 0

3 years ago

Help meeeee doooo thissss

lesya692 [45]

Answer:

B: x=-5 and x=4

Step-by-step explanation:

f(x)= x^2 + x -20

To find the zeros we replace f(x) with 0

$0= x^2 + x -20$

now we solve for x

LEts factor x^2 + x -20

sum is +1 and product is -20

5 * (-4)= -20

5 + (-4) = +1

x^2 + 5x - 4x - 20 = 0

(x^2 + 5x)+(- 4x - 20) = 0

x(x+5) -4 (x+5) = 0

(x-4)(x+5)=0

Set each factor =0 and solve for x

x- 4 =0 so x=4

x+5 =0 so x=-5

x=-5 and x=4

3 0

3 years ago

Please help

trapecia [35]

Between emma and dave, we have 7/10ths of the money

3x + 2x = 7/10

5x = 7/10

x=(7/10)/5

x= 14/100

Dave will get 28/100 or 7/25ths

Check : 30/100 Colin
42/100 Emma
+ 28/100 Dave
-----------
100/00

7 0

3 years ago

(05.01)

GarryVolchara [31]

3x - 6y = 72...reduce by dividing everything by 3
x - 2y = 24...notice how this is exactly the same as the other equation....means that the lines are the same line...meaning infinite solutions

3 0

3 years ago

Read 2 more answers

Evaluate the following integral (Calculus 2) Please show step by step explanation!

barxatty [35]

Answer:

$\dfrac{1}{2} \left(25 \arcsin \left(\dfrac{x}{5}\right) -x\sqrt{25-x^2}\right) + \text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{x^2}{\sqrt{25-x^2}}\:\:\text{d}x$

Rewrite 25 as 5²:

$\implies \displaystyle \int \dfrac{x^2}{\sqrt{5^2-x^2}}\:\:\text{d}x$

Integration by substitution

$\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=5 \sin \theta$

$\begin{aligned}\implies \sqrt{5^2-x^2} & =\sqrt{5^2-(5 \sin \theta)^2}\\ & = \sqrt{25-25 \sin^2 \theta}\\ & = \sqrt{25(1-\sin^2 \theta)}\\ & = \sqrt{25 \cos^2 \theta}\\ & = 5 \cos \theta\end{aligned}$

Find the derivative of x and rewrite it so that dx is on its own:

$\implies \dfrac{\text{d}x}{\text{d}\theta}=5 \cos \theta$

$\implies \text{d}x=5 \cos \theta\:\:\text{d}\theta$

Substitute everything into the original integral:

$\begin{aligned}\displaystyle \int \dfrac{x^2}{\sqrt{5^2-x^2}}\:\:\text{d}x & = \int \dfrac{25 \sin^2 \theta}{5 \cos \theta}\:\:5 \cos \theta\:\:\text{d}\theta \\\\ & = \int 25 \sin^2 \theta\end{aligned}$

Take out the constant:

$\implies \displaystyle 25 \int \sin^2 \theta\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad \cos (2 \theta)=1 - 2 \sin^2 \theta$

$\implies \displaystyle 25 \int \dfrac{1}{2}(1-\cos 2 \theta)\:\:\text{d}\theta$

$\implies \displaystyle \dfrac{25}{2} \int (1-\cos 2 \theta)\:\:\text{d}\theta$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\cos kx$}\\\\$\displaystyle \int \cos kx\:\text{d}x=\dfrac{1}{k} \sin kx\:\:(+\text{C})$\end{minipage}}$

$\begin{aligned} \implies \displaystyle \dfrac{25}{2} \int (1-\cos 2 \theta)\:\:\text{d}\theta & =\dfrac{25}{2}\left[\theta-\dfrac{1}{2} \sin 2\theta \right]\:+\text{C}\\\\ & = \dfrac{25}{2} \theta-\dfrac{25}{4}\sin 2\theta + \text{C}\end{aligned}$

$\textsf{Use the trigonometric identity}: \quad \sin (2 \theta)= 2 \sin \theta \cos \theta$

$\implies \dfrac{25}{2} \theta-\dfrac{25}{4}(2 \sin \theta \cos \theta) + \text{C}$

$\implies \dfrac{25}{2} \theta-\dfrac{25}{2}\sin \theta \cos \theta + \text{C}$

$\implies \dfrac{25}{2} \theta-\dfrac{5}{2}\sin \theta \cdot 5 \cos \theta + \text{C}$

$\textsf{Substitute back in } \sin \theta=\dfrac{x}{5} \textsf{ and }5 \cos \theta = \sqrt{25-x^2}:$

$\implies \dfrac{25}{2} \theta-\dfrac{5}{2}\cdot \dfrac{x}{5} \cdot \sqrt{25-x^2} + \text{C}$

$\implies \dfrac{25}{2} \theta-\dfrac{1}{2}x\sqrt{25-x^2} + \text{C}$

$\textsf{Substitute back in } \theta=\arcsin \left(\dfrac{x}{5}\right) :$

$\implies \dfrac{25}{2} \arcsin \left(\dfrac{x}{5}\right) -\dfrac{1}{2}x\sqrt{25-x^2} + \text{C}$

Take out the common factor 1/2:

$\implies \dfrac{1}{2} \left(25 \arcsin \left(\dfrac{x}{5}\right) -x\sqrt{25-x^2}\right) + \text{C}$

Learn more about integration by trigonometric substitution here:

brainly.com/question/28157322

6 0

2 years ago