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Brums [2.3K]
3 years ago
9

What goes in the blanks? (a2 +___+1)−(___+5a+___)=4a2 −2a+7

Mathematics
2 answers:
andrey2020 [161]3 years ago
6 0

Answer:

4a2 + 4a + 1 = (2a +1)^2 = (2a + 1)(2a + 1)

4 − 4a + a2  = (2 - a)^2 = (2 − a)(2 − a)

4a2 − 4a + 1 = (2a -1)^2 = (2a − 1)(2a − 1)

4 + 4a + a2 = (2 + a)^2 = (2 + a)(2 + a)

Step-by-step explanation:

oksian1 [2.3K]3 years ago
4 0

Answer:

  • 3a, -3a², - 6

Step-by-step explanation:

<u>The left side of the equation:</u>

  • a² - 5a + 1

<u>The right side:</u>

  • 4a² - 2a + 7

<u>The difference:</u>

  • - 3a² + 3a - 6

<u>The equation is:</u>

  • (a² + 3a + 1) - (-3a² + 5a - 6) = 4a² - 2a + 7

<u>The blanks:</u>

  • 3a, -3a², - 6
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The statement " <span>y </span>varies directly as <span>x </span>," means that when <span>x </span>increases,<span>y </span>increases by the same factor. In other words, <span>y </span>and <span>x </span>always have the same ratio:

<span><span> = k</span>   </span>

<span>where </span><span>k </span>is the constant of variation.
<span>We can also express the relationship between </span><span>x </span><span>and </span><span>y </span>as:

<span><span>y = kx</span>   </span>

<span>where </span><span>k </span>is the constant of variation.

Since <span>k </span>is constant (the same for every point), we can find <span>k </span>when given any point by dividing the y-coordinate by the x-coordinate. For example, if <span>y </span>varies directly as <span>x </span>, and <span>y = 6</span> when <span>x = 2</span> , the constant of variation is <span>k =  = 3</span> . Thus, the equation describing this direct variation is <span>y = 3x </span>.

Example 1: If <span>y </span>varies directly as <span>x </span>, and <span>x = 12</span> when <span>y = 9</span> , what is the equation that describes this direct variation?

<span>k =  =  </span>
<span>y =  x</span>

Example 2: If <span>y </span>varies directly as <span>x </span>, and the constant of variation is <span>k =  </span>, what is <span>y </span>when <span>x = 9</span> ?

<span>y =  x = (9) = 15</span>

As previously stated, <span>k </span>is constant for every point; i.e., the ratio between the <span>y </span>-coordinate of a point and the <span>x </span>-coordinate of a point is constant. Thus, given any two points <span>(x 1, y 1)</span> and <span>(x 2, y 2)</span> that satisfy the equation, <span> = k </span>and <span> = k </span>. Consequently, <span> =  </span>for any two points that satisfy the equation.

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kolbaska11 [484]

\qquad\qquad\huge\underline{{\sf Answer}}♨

Let's check I the given pairs of triangles are congruent or not ~

<h3>problem 1 </h3>

\qquad \sf  \dashrightarrow \:ABC \cong FED

These triangles are congruent by SSS congruency, as it's three sides are equal.

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 2 </h3>

\qquad \sf  \dashrightarrow \:LMN \not \cong PQO

・ .━━━━━━━†━━━━━━━━━.・

These triangles are not congruent because they don't fulfill any congruency criteria.

<h3>problem 3 </h3>

\qquad \sf  \dashrightarrow \: XVW \cong YZA

These triangles are congruent by AAS congruency criteria, since two angles and one of the side is common.

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 4 </h3>

\qquad \sf  \dashrightarrow \: GHI \not \cong IJG

These triangles are not congruent because they don't fulfill any criteria.

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 5</h3>

\qquad \sf  \dashrightarrow \:  PQR\cong  QPO

These triangles are congruent by SSS congruency, as it's two sides are equal and one side is common.

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 6</h3>

\qquad \sf  \dashrightarrow \:  JIK \cong  HIG

These triangles are congruent by SAS congruency, since two sides and Angle between them are equal.

・ .━━━━━━━†━━━━━━━━━.・

<h3>Problem 7</h3>

These given triangles are congruent by SSS congruency, because they have two equal sides and one common side in there.

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 8 </h3>

\qquad \sf  \dashrightarrow \:  BCD  \cong  FED

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・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 9</h3>

\qquad \sf  \dashrightarrow \:    LKM \cong  LNM

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5 0
2 years ago
1
pantera1 [17]

Answer:

The probability of winning directly is, as you calculated, 8/36, and the probability of losing directly is (1+2+1)/36=4/36.

For the remaining cases, you need to sum over all remaining rolls. Let p be the probability of rolling your initial roll, and q=6/36=1/6 the probability of rolling a 7. Then the probability of rolling your initial roll before rolling a 7 is p/(p+q), and the probability of rolling a 7 before rolling your initial roll is q/(p+q). Thus, taking into account the probability of initially rolling that roll, each roll that doesn't win or lose directly yields a contribution p2/(p+q) to your winning probability.

For p=5/36, that's

(536)25+636=2511⋅36,

and likewise 16/(10⋅36) and 9/(9⋅36) for p=4/36 and p=3/36, respectively. Each of those cases occurs twice (once above 7 and once below), so your overall winning probability is

836+236(2511+1610+99)=244495=12−7990≈12−0.007.

Step-by-step explanation:

Suppose you throw a 4 and let p(4) your winning probability. At your next roll you have a probability 3/36 of winning (you throw a 4), a probability 6/36 of losing (you throw a 7) and a probability 27/36 of repeating the whole process anew (you throw any other number). Then:

p(4)=336+2736p(4),so thatp(4)=13.

Repeat this reasoning for the other outcomes and then compute the total probability of winning as:

ptot=836+336p(4)+436p(5)+…

7 0
2 years ago
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