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Yuki888 [10]
3 years ago
8

Please answer the question in the picture please its urgent 200 pts plus brainly!!!!

Mathematics
1 answer:
OverLord2011 [107]3 years ago
6 0

Answer:

The answers B.

Step-by-step explanation:

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I'm so lost. I have no clue.
tiny-mole [99]
I supppose the answer is F(X)=X²+4, you can use the vertex to confirm.
7 0
3 years ago
Read 2 more answers
A certain number was multiplied by 7. 4 was taken away from the product. Finally, that difference was then divided by 9, resulti
marin [14]

Answer:

7

Step-by-step explanation:

To solve this, we have to work backwards. If you divide something by 9, and it equals 5, to figure out the number, we need to multiply by 9. 5*9= 45. If four was taken away, we have to add it to 45, which is 49. If a number *7 is 49, then we have to divide by 7 to get the answer. 49/7=7  

7 0
3 years ago
Sales of 10000 increased by 65% each year write a function that represents the situation
erastovalidia [21]

Answer: y=10000(1.65)^x


Step-by-step explanation:

Given: The sales of 10000 increased by 65% each year .

Let x be the number of years.

The rate of growth=65%= 0.65

The exponential growth function is given by

y=A(1+r)^x

, where A is initial amount and r is the rate of growth in decimal.

Then the increase in the sales id given by :-

y=10000(1+0.65)^x\\\Rightarrow\ y=10000(1.65)^x

6 0
3 years ago
Anyone knows how to do questions 7 and 8? 15 pts!!
Oksi-84 [34.3K]
7) Certainly there is a typo in the statement, just see that the expression of item (ii) is different from that of item (i). Probably the correct expression is: 2x^2-4x+5. With this consideration, we can continue.

(i) Let E the expression that we are analyzing:

E=2x^2-4x+5\\\\ E=2x^2-4x+2-2+5\\\\ E=2(x^2-2x+1)-2+5\\\\ E=2(x-1)^2+3

Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.

(ii) Manipulating the expression:

2x^2+5=4x\\\\ 2x^2-4x+5=0

So, it's the case when E=0. However, E is always a positive number. Then, there is no real number x that satisfies the expression.

8) Let E the expression that we want to calculate:

E=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)+1\\\\ E-1=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)

Multiplying by (2-1) in the both sides:

(2-1)(E-1)=(2-1)(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2-1)(2+1)}_{2^2-1}(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2^2-1)(2^2+1)}_{2^4-1}(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ ... Repeating the process, we obtain: ...\\\\ E-1=(2^{32}-1)(2^{32}+1)\\\\ E-1=2^{64}-1\\\\ \boxed{E=2^{64}}
3 0
3 years ago
Read 2 more answers
Round 0.42 to the nearst tenth
Archy [21]
0.4 because if it’s five or more, add one more. If it’s four or less, let it rest.
4 0
3 years ago
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