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Vitek1552 [10]
3 years ago
15

2. Solve for the x. 8 16 7

Mathematics
1 answer:
olasank [31]3 years ago
3 0

Answer:

Theres no equation here?

Step-by-step explanation:

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Given the functions below, find f(x) - g(x) f(x) = 2x + 5 g(x) = x^2 - 3x + 1
Inessa [10]

Answer:

One of these

Step-by-step explanation:

6 0
3 years ago
The cost of a school banquet is $95 plus $15 for each person attending. Right and equation that gives total cost as a function o
maks197457 [2]
15P+95=C P=each person attending C=total cost
5 0
4 years ago
What percent of 150 is 350
DIA [1.3K]
p\% \ of \ 150=350 \\\\ \frac{p}{100}*150=350 \\\\ \frac{p}{2}*3=350 \\\\ \frac{p}{2}=\frac{350}{3} \\\\ p=\frac{2*350}{3}=\frac{700}{3} \\\\ \boxed{p=233.33 \%}
7 0
4 years ago
if a and b are the two odd positive integers such that a>b then one of the two numbers ( a+b) /2 and (a-b) /2is odd and even
Tamiku [17]

If a and b are odd positive integers, they are one more than a non-negative even number, i.e. there exists m,n \in \mathbb{N} such that

a = 2m+1,\quad b = 2n+1

So, the first expression become

\cfrac{a+b}{2} = \cfrac{2m+1+2n+1}{2} = \cfrac{2m+2n+2}{2} = m+n+1

Similarly, we have

\cfrac{a-b}{2} = \cfrac{2m+1-2n-1}{2} = \cfrac{2m-2n}{2} = m-n

Now, the parity of these expressions depend on those of m and n. We have four cases:

If both m and n are even:

m+n+1 is odd, since m+n is even, while m-n is even

If one of the two is odd and the other is even:

m+n+1 is even, since m+n is odd, while m-n is odd

If both are odd:

m+n+1 is odd, since m+n is even, while m-n is even

So, in all cases, one between (a+b)/2 and (a-b)/2 is odd, and the other is even.

7 0
3 years ago
PLEASE, PLEASE HELP!!! WILL MARK BRAINLIEST!!!
Irina18 [472]

Rewrite the original equation as:

Log(4x^2) - log(3yz)

Rewrite log(4x^2) as log(4) + log(x^2)

Rewrite log(4) as 2log(2)

Rewrite log(x^2) as 2log(x)

Separate log(3yz) into 3 logs: log(3), log(y) and log(z)

Now combine them to get:

2log(2) + 2log(x) - log(3) - log(y) - log(z)

5 0
4 years ago
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