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kupik [55]
3 years ago
14

Will is w years old Ben is 3 years older Write an expression, in terms of w, for bens age

Mathematics
1 answer:
Westkost [7]3 years ago
7 0
Ben is w+3 years old.
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Please answer quick!
inessss [21]
The answer should be the 3rd one
5 0
3 years ago
A 99 ft length of wire is to be cut into 2 pieces, so that the longer piece is eight feet longer than six times the shorter piec
Serhud [2]

here's the solution,

let the length of shorter peice = x

then the lenght of longer peice = 6x + 8

now, we know that the sum of length of peices measures 99 feet,

So,

=》x + 6x + 8 = 99

=》7x = 91

=》x = 13

shorter piece length = 13 feet

longer peice length = 13 × 6 + 8 = 78 + 8 = 86 feet

5 0
3 years ago
Please explain the how you got the answers thanks asap
stich3 [128]

Answer:

a) 90 stamps

b) 108 stamps

c) 333 stamps

Step-by-step explanation:

Whenever you have ratios, just treat them like you would a fraction! For example, a ratio of 1:2 can also look like 1/2!

In this context, you have a ratio of 1:1.5 that represents the ratio of Canadian stamps to stamps from the rest of the world. You can set up two fractions and set them equal to each other in order to solve for the unknown number of Canadian stamps. 1/1.5 is representative of Canada/rest of world. So is x/135, because you are solving for the actual number of Canadian stamps and you already know how many stamps you have from the rest of the world. Set 1/1.5 equal to x/135, and solve for x by cross multiplying. You'll end up with 90.

Solve using the same method for the US! This will look like 1.2/1.5 = x/135. Solve for x, and get 108!

Now, simply add all your stamps together: 90 + 108 + 135. This gets you a total of 333 stamps!

5 0
3 years ago
AD is the perpendicular bisector of BC. BD = 5x - 10 and DC = 3x + 10. Determine the value of x.
Nookie1986 [14]

Answer: x= 10

Step-by-step explanation:

6 0
3 years ago
Find the derivative of<br> <img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%5Cfrac%7B6%7D%7Bx%7D%20" id="TexFormula1" titl
aev [14]
f'(x_0)=\lim\limits_{h\to0}\dfrac{f(x_0+h)-f(h)}{h}=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\\f(x)=\dfrac{6}{x};\ x_0=2\\\\subtitute\\\\f'(2)=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-\frac{6}{2}}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6}{x}-3}{x-2}=\lim\limits_{x\to2}\dfrac{\frac{6-3x}{x}}{x-2}=\lim\limits_{x\to 2}\dfrac{6-3x}{x(x-2)}\\\\=\lim\limits_{x\to2}\dfrac{-3(x-2)}{x(x-2)}=\lim\limits_{x\to2}\dfrac{-3}{x}=-\dfrac{3}{2}=-1.5\\\\\\An swer:\boxed{f'(2)=-1.5}
8 0
3 years ago
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