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2 years ago

Hello, I know it is kinda late but here's a question.

Mathematics

1 answer:

zaharov [31]2 years ago

8 0

The solution of the linear equation is a = 11/4.

<h3></h3><h3>How to solve the linear equation?</h3>

Here we have the linear equation:

9 = 4a - 2

We want to solve this for a, so we need to isolate a on one side of the equation.

First, we can add 2 on both sides to get:

9 + 2 = 4a

11 = 4a

Now we can divide both sides by 4 to get:

11/4 = a

The solution of the linear equation is a = 11/4.

If you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

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3. How many 1/8's are in 1 1/4?

Murrr4er [49]

Answer:

Step-by-step explanation:

There are two 1/8 in one 1/4. There are eight 1/8 in 1. 8/8 + 1/8 + 1/8 = ten 1/8.

3 0

3 years ago

Filipemarcosdesouza

Margaret [11]

Answer:

noted, thanks for the note

7 0

3 years ago

Read 2 more answers

If the perimeter is 102, how long is the longest side?

aalyn [17]

25.5
102/4=25.5

Not really sure though

8 0

3 years ago

Solve the following proportion for x. 3/13 = x/5 Round your answer to the nearest tenth.

anastassius [24]

To solve, isolate the x. Cross multiply

3/13(13)(5) = x/5(5)(13)

3(5) = (13)x

Simplify

13x = 15

Isolate the x. Divide by 13.

13x/13 = 15/13

x = 15/13

x = 1.2

1.2 is your answer

hope this helps

8 0

3 years ago

Hi, can someone please help, and possibly explain the answer please.

mr_godi [17]

Answer:

$x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}$

Step-by-step explanation:

Quadratic formula: $x=\frac{-b\±\sqrt{b^2-4ac} }{2a}$ when the equation is $0=ax^2+bx+c$

The given equation is $1=-2x+3x^2+1$ . Let's first arrange this so its format looks like $y=ax^2+bx+c$ :

$1=-2x+3x^2+1$

$1=3x^2-2x+1$

Subtract 1 from both sides of the equation

$1-1=3x^2-2x+1-1\\0=3x^2-2x+0$

Now, we can easily identify 3 as a, -2 as b and 0 as c. Plug these into the quadratic formula:

$x=\frac{-b\±\sqrt{b^2-4ac} }{2a}\\x=\frac{-(-2)\±\sqrt{(-2)^2-4(3)(0)} }{2(3)}$

I hope this helps!

8 0

3 years ago