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slava [35]
3 years ago
11

WILLL GIVE BRAINLIEST + 50 POINTS FOR HELP!!!!PLZZZZZZ IM TIMED

Mathematics
2 answers:
andrew-mc [135]3 years ago
7 0
First. y=2x
second d is correct
third is a
Ilia_Sergeevich [38]3 years ago
6 0
Idk but i want the pints good luck bby
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Find two consecutive odd integers whose product is 323
Harrizon [31]
Use a let statement 
first
let x and x + 2 be the number so you write it like this
<u>let x = the first consecutive integer
</u><u>let x + 2 = the second consecutive integer
</u>
second
x(x+2)=323
x^2 + 2x = 323
        -323  -323
x^2 + 2x -323 = 0

third
try to factor -323 so it is 19 and -17
(x + 19) (x - 17) = 0
x = 19 
x = -17

hope this help

4 0
3 years ago
Finn is walking on a treadmill at a constant pace for 30 minutes. She has programmed the treadmill for a 2-mile walk. The displa
Ymorist [56]

30*2=15miles per min. x*y=d.

4 0
3 years ago
Read 2 more answers
What is the probability of flipping two coins at the same time and getting heads on both coins?
Romashka-Z-Leto [24]
According to me the probability is 1/4, because there are 4 possible outcomes when two coins are flipped - TT, TH, HT, HH. 

<span>Also, would it matter if the coins are flipped one after other rather than together</span>
5 0
3 years ago
Read 2 more answers
Let P(x) denote the statement "2x+5 &gt; 10." Which of the following is true?
ruslelena [56]

Answer: P(3) is True

Step-by-step explanation:

The given statement is an inequality denoted as P(x). To find out which of the options is true you have to evaluate each given value of X in the inequality and perform the arithmetic operations, then you have to see if the expression makes sense.

For P(0): Replace X=0 in 2x+5>10

2(0)+5>10

0+5>10

5>10 is false because 5 is not greater than 10

For P(3): Replace X=3 in 2x+5>10

2(3)+5>10

6+5>10

11>10 is true because 11 is greater than 10

For P(2): Replace X=2 in 2x+5>10

2(2)+5>10

4+5>10

9>10 is false

For P(1): Replace X=1 in 2x+5>10

2(1)+5>10

2+5>10

7>10 is false

4 0
3 years ago
Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator
SpyIntel [72]
\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}
\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)

When x=3, you're left with

147=49a_1\implies a_1=\dfrac{147}{49}=3

When x=\sqrt2 or x=-\sqrt2, you're left with

\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}

Adding the two equations together gives -10=2a_5, or a_5=-5. Subtracting them gives 2\sqrt2=2\sqrt2a_4, a_4=1.

Now, you have

5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)
5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)
2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that a_2x^4=2x^4 and a_3(-3)(-2)=6a_3=-6. These alone tell you that you must have a_2=2 and a_3=-1.

So the partial fraction decomposition is

\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}
7 0
3 years ago
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