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lara [203]
3 years ago
10

This is a double question. Part 1: 5(2x - 4) = 30. Solve for x Part 2: 5x + 3 - x + 2 = 41. Solve for x. I will mark Brainliest

to whoever solves both parts as well as explains it.
Mathematics
2 answers:
SOVA2 [1]3 years ago
4 0
PART 1:
5(2x-4)=30
Multiply both 2x and -4 by 5

10x-20=30
Then add 20 to cancel it out, but once u do something to that side u must do it to the other side. So add 20 to 30 which is 50.

10x=50

10x/10= 50/10
Divide both side by 10

X=5



PART 2:
5x + 3 - x + 2 = 41

4x + 5 = 41
Combine the like terms

4x=36
Then subtract 5 from both sides to cancel to get 4x by itself

4x/4=36/4
Then divide both sides by 4

X=9
Ainat [17]3 years ago
3 0

Answer:

see below

Step-by-step explanation:

Part 1: 5(2x - 4) = 30.

Distribute

10x -20 = 30

Add 20 to each side

10x-20+20 =30+20

10x = 50

Divide by 10

10x/10 = 50/10

x =5

Part 2: 5x + 3 - x + 2 = 41

Combine like terms

4x +5 = 41

Subtract 5 from each side

4x+5-5 = 41-5

4x = 36

Divide by 4

4x/4 = 36/4

x = 9

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7 0
3 years ago
Which best describes the composition of two functions
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The composition of 2 functions is B
3 0
3 years ago
2x – 3y = 18<br> Solve for y.
Nataly [62]

Answer:

y= -6+2/3x

Step-by-step explanation:

2x-3y=18

Minus the 2x from both sides

-3y=18-2x

Then divide both sides by -1

-3y÷-1=18÷-1 -2x÷-1

Lastly divide both sides by 3

3y÷3= -18÷3+2x÷3

7 0
3 years ago
You want to get from a point A on the straight shore of the beach to a buoy which is 54 meters out in the water from a point B o
anyanavicka [17]

Answer:

x =\dfrac{45 \sqrt{6}}{ 2}

Step-by-step explanation:

From the given information:

The diagrammatic interpretation of what the question is all about can be seen in the diagram attached below.

Now, let V(x) be the time needed for the runner to reach the buoy;

∴ We can say that,

\mathtt{V(x) = \dfrac{70-x}{7}+\dfrac{\sqrt{54^2+x^2}}{5}}

In order to estimate the point along the shore, x meters from B, the runner should  stop running and start swimming if he want to reach the buoy in the least time possible, then we need to differentiate the function of V(x) and relate it to zero.

i.e

The differential of V(x) = V'(x) =0

=\dfrac{d}{dx}\begin {bmatrix} \dfrac{70-x}{7} + \dfrac{\sqrt{54^2+x^2}}{5} \end {bmatrix}= 0

-\dfrac{1}{7}+ \dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}=0

\dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}

\dfrac{5x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}

\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{\dfrac{7}{5}}

\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{5}{7}

squaring both sides; we get

\dfrac{x^2}{54^2+x^2}= \dfrac{5^2}{7^2}

\dfrac{x^2}{54^2+x^2}= \dfrac{25}{49}

By cross multiplying; we get

49x^2 = 25(54^2+x^2)

49x^2 = 25 \times 54^2+ 25x^2

49x^2-25x^2 = 25 \times 54^2

24x^2 = 25 \times 54^2

x^2 = \dfrac{25 \times 54^2}{24}

x =\sqrt{ \dfrac{25 \times 54^2}{24}}

x =\dfrac{5 \times 54}{\sqrt{24}}

x =\dfrac{270}{\sqrt{4 \times 6}}

x =\dfrac{45 \times 6}{ 2 \sqrt{ 6}}

x =\dfrac{45 \sqrt{6}}{ 2}

8 0
3 years ago
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