Ask question

Ask question

All categories

2 years ago

11

The cost of 12 oranges and 7 coples is 55.36. Eight oranges and apples cost $3.68. Find the cost

1 answer:

victus00 [196]2 years ago

4 0

Hello there i hope you are having a good day :) Your question : The cost of 12 oranges and 7 couples is 55.36. Eight oranges and apples cost $3.68. Find the cost of each : Firstly you would do oranges = O Apples = A. Then 12 oranges + 7 apples = 5.36 12o + 7A = 5.36 then you would take away from both sides so -12o Take away -12o = 7o = 5.36 - 12A now you would divide this by 7 so 7 cancels each out so you are left with 5.36/ 7 = 0.76 Then it would equal 0.76 - 12o/7 This would equal the amount of apples to be $0.76 each now for the oranges you would do the same that you work out for the apples.

Hopefully that help ya :)

You might be interested in

What is 4(x−2+y)? I need help.

stepladder [879]

4x+4y-8

Hope this helps!

7 0

2 years ago

Let f(x)=3x+5 and g(x)=x^2 find (f-g)(x)

dlinn [17]

Answer:

(f - g)(x) = -x² + 3x + 5

General Formulas and Concepts:

<u>Pre-Algebra</u>

Equality Properties

<u>Algebra I</u>

Function Notation
Combining Like Terms

Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = 3x + 5

g(x) = x²

(f - g)(x) is f(x) - g(x)

<u>Step 2: Find (f - g)(x)</u>

Substitute: (f - g)(x) = 3x + 5 - x²
Rewrite: (f - g)(x) = -x² + 3x + 5

7 0

2 years ago

Please go to my profile and answer my last question

Alex Ar [27]

Answer:

ok

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Graham is graphing the inequality x greater-than 4. He draws a number line and shades over the numbers that are greater than 4.

schepotkina [342]

Answer:

Draw an open circle at 4

Step-by-step explanation:

8 0

2 years ago

Which of the following expressions is a factor of the polynomial x 2 +3/2x-1

Elodia [21]

Answer:

$\large \boxed{\sf \ \ \ (x+2)(x-\dfrac{1}{2}) \ \ \ }$

Step-by-step explanation:

Hello,

let's solve

$x^2+\dfrac{3}{2}x-1=0\\\\ 2x^2+3x-2=0 \ \text{multiply by 2}\\\\\\$

$\Delta=b^2-4ac=9+4*2*2=9+16=25$

There are two solutions

$x_1=\dfrac{-3-\sqrt{25}}{4}\\\\x_1=\dfrac{-3-5}{4}\\\\x_1=\dfrac{-8}{4}\\\\\boxed{x_1=-2}$

And

$x_2=\dfrac{-3+\sqrt{25}}{4}\\\\x_2=\dfrac{-3+5}{4}\\\\x_2=\dfrac{2}{4}\\\\\boxed{x_2=\dfrac{1}{2}}$

So we can write

$x^2+\dfrac{3}{2}x-1\\\\=(x-x_1)(x-x_2)\\\\=(x+2)(x-\dfrac{1}{2})$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

5 0

2 years ago