Answer:
5m - 4n
5(5) - 4(3) ( As m= 5 and n= 3)
25 - 12
= 13
Y=-3x+2 bc the line is going down so it’s negative and the y intercept is 2
The perimeter would be (24x - 40)/(x^2 - 4x).
In order to find this, first double the length and width as you would to find any perimeter.
7/(x - 4) * 2 = 14/(x - 4)
5/x * 2 = 10/x
Now to add those together, we need to give them common denominators. In order to do that with the first one, we need to multiply by x/x
14/(x - 14) * x/x = 14x/(x^2 - 14x)
Then we can do the same with the second part by multiplying by (x - 4)/(x - 4)
10/x * (x - 4)/(x - 4) = (10x - 40)/(x^2 - 14x)
Now we can add the two together
14x/(x^2 - 14x) + (10x - 40)/(x^2 - 14x) = (24x - 40)/(x^2 - 14x)
Answer:
16.75 g
Step-by-step explanation:
Orange's weight = 67 g
Strawberry's weight
= ¼ of the weight of orange
= ¼ × weight of orange
= ¼ × 67 g
= 67 g/4
= 16.75 g
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4.
a. $100
b. $360
The formula to find the answers to these kinds of problems is p * r * t. The rate is always a decimal, so if the rate is 8%, you multiply by 0.08. If the rate is 12%, you multiply by 0.12, and so on.
5.
a. 12.8
b. 31 1/6
I got these answers by doing the distributive property. Replace each variable with the number it represents, multiply each by two and add them up to get these answers. For the fraction one, you can set up each fraction by finding the LCD before multiplying by 2, or you can find a fraction calculator online and add the numbers up.
6.
a. x = 20
b. x = 3
c. x = 13
d. x = 0
e. x = 4
f. x = 7
g. x = -3
I solved this problem by doing the distributive property and doing these kinds of problems step by step. It’s long to explain, but hopefully your teacher has good notes for you to look back on over these types of questions. If there is only a negative sign in front of the distributive property equation, then you basically distribute -1 to each number. In these long equations, combine like terms. You must do these kinds of questions right, because it’s easy to mess up and get wrong answers if you don’t do the right steps in order.
