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2 years ago

3(2x-1)=5x+3+7x I do t get this cause I’m stupid

Mathematics

2 answers:

Tcecarenko [31]2 years ago

8 0

Answer:

x=-1

Step-by-step explanation:

first you multiply 3 by whats in the parenthesis (you would get 6x-3)

then on the other side you combine like terms (add 7x plus 5x=12x)

Then you would isolate X to one side by subtracting 6x from 12x

Then you would subtract 3 from the side with 12x

By now your equation would be -6=6x

so you would divide both side by negative six to get x=-1

pishuonlain [190]2 years ago

6 0

Answer:

x=-1

Step-by-step explanation:

3(2X-1)=5x+3+7x Distribute the 3 into the parenthesis

6x-3=5x+3+7x Isolate the x on one side

-6x -6x

-3=5x+3+7x-6x

-3 -3

-6=5x+7x-6x Add all the x's

-6=6x Divide 6 to isolate x

(-6)/6=(6x)/6

-1=x

Now you can check if it's right by inputting -1 for x and seeing if the equations equal each other

3(2(-1)-1)=5(-1)+3+7(-1)

3(-3)=-5+3-7

-9=-9 They equal each other so x=-1 is correct

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Donna sells pencils for $2 and pens $3. She sold 11 items and made $30. How many pens did she sell ?

finlep [7]

Donna sold 3 pencils and 8 pens solved using simultaneous equation

What is a simultaneous equation?

Simultaneous equation refers to related equations that are solved at the same to derive answers to the unknown variables

Let assume that X pencils were sold and Y pens were sold

The total revenue is the sum of the prices multiplied by the quantities sold.

2X+3Y=30 (equation 1)

The sum of X and Y sold is 11

X+Y=11 (equation 2)

multiply the second equation by 2

2*(X+Y=11)

2X+2Y=22 (equation 3)

subtract equation 3 from 1

(2X+3Y=30)-(2X+2Y=22)

Y=8

Substitute for Y in equation 2

X+8=11

X=11-8

X=3

Which means he sold 3 pencils and 8 pens.

Find details on simultaneous equation in the link below:

brainly.com/question/20627346

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4 0

1 year ago

Find the unknown side length. Write your answer in simplest radical form.

Eva8 [605]

Answer: 1. x= 20 units, 2. x= $5\sqrt{3}$ units 3. x= $4\sqrt{2}$ units.

Step-by-step explanation:

In a right triangle,

$Hypotenuse^2= Perpendicular^2+Base^2$ (Pythagoras theorem)

1. Hypotenuse = x , Base = 16, Perpendicular = 12

$\Rightarrow\ x^2 =12^2+16^2\\\\ \Rightarrow\ x^2=144+256\\\\ \Rightarrow\ x^2=400\\\\ \Rightarrow\ x=\sqrt{400}=20$

2. Hypotenuse = 14 , Base = 11, Perpendicular = x

$\Rightarrow\ 14^2=x^2+11^2\\\\\Rightarrow\ 196=x^2+121\\\\\Rightarrow\ x^2=196-121\\\\\Rightarrow\ x^2 = 75\\\\\Rightarrow\ x=\sqrt{75}=\sqrt{25\times3}=5\sqrt{3}$

3. Hypotenuse = 9 , Base = x, Perpendicular =7

$\Rightarrow\ 9^2=x^2+7^2\\\\\Rightarrow\ 81= x^2+49\\\\\Rightarrow\ x^2=81-49\\\\\Rightarrow\ x^2=32\\\\\Rightarrow\ x=\sqrt{32}=\sqrt{16\times2}\\\\\Rightarrow\ x=4\sqrt{2}$