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RideAnS [48]
3 years ago
15

(09.02 MC) Which of the following tables shows the correct steps to transform x2 + 10x + 24 = 0 into the form (x − p)2 = q? [p a

nd q are integers] (5 points) Select one: a. Step 1 x2 + 10x + 24 − 1 = 0 − 1 Step 2 x2 + 10x + 23 = −1 Step 3 (x + 5)2 = −1 b. Step 1 x2 + 10x + 24 − 2 = 0 − 2 Step 2 x2 + 10x + 22 = −2 Step 3 (x + 5)2 = −2 c. Step 1 x2 + 10x + 24 + 2 = 0 + 2 Step 2 x2 + 10x + 26 = 2 Step 3 (x + 5)2 = 2 d. Step 1 x2 + 10x + 24 + 1 = 0 + 1 Step 2 x2 + 10x + 25 = 1 Step 3 (x + 5)2 = 1
Mathematics
2 answers:
Virty [35]3 years ago
8 0
What is this way too much for my brain
Vladimir79 [104]3 years ago
4 0

Answer: step 2

Step-by-step explanation:

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-15x + 5 = -10x + 25 solve for x
tatuchka [14]

Answer:

Move all terms containing x to the left side of the equation.

−5x+5=25

Move all terms not containing x to the right side of the equation.

−5x=20

Divide each term by −5 and simplify.

x=−4

Step-by-step explanation:

6 0
2 years ago
Write a number in which the value of the 3 is ten times greater than the value of the 3 in 135, 864
Aleonysh [2.5K]

Answer:

345,456

Step-by-step explanation:

The value of 3 in 135,864 is 30,000.

We want to write a number that is ten times 30,000.

We could write any other number in which the value of 3 is 300,000.

There are infinitely many numbers that we can write.

Some examples are:

345,456

1,315,445

5,354,456

and so on and so forth.

8 0
3 years ago
Triangles ABC and DEF have the following characteristics B and Eare right angles A = D BC = EF which congruence theorem can be u
kvasek [131]
AAS Angles A and D are one pair of congruent angles, then angles B and E are congruent because they are both right angles, following that you have line BC as congruent to EF so that is an angle, an angle, and a side so AAS

Read more on Brainly.com - brainly.com/question/7297105#readmore
6 0
2 years ago
8/x-5 - 9/x-4 = 5/x^2-9x+20
adelina 88 [10]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2752942

_______________


Solve the equation:

\mathsf{\dfrac{8}{x-5}-\dfrac{9}{x-4}=\dfrac{5}{x^2-9x+20}\qquad\qquad(x\ne 5~and~x\ne 4)}


Reduce the fractions at the left side so that they have the same denominator:

\mathsf{\dfrac{8(x-4)}{(x-5)(x-4)}-\dfrac{9(x-5)}{(x-4)(x-5)}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-4x-5x+20}-\dfrac{9x-45}{x^2-4x-5x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-9x+20}-\dfrac{9x-45}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32-(9x-45)}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}


Numerators must be equal:

\mathsf{8x-32-(9x-45)=5}\\\\
\mathsf{8x-32-9x+45=5}\\\\
\mathsf{8x-9x=5+32-45}\\\\
\mathsf{-x=-8}\\\\
\mathsf{x=8}\quad\longleftarrow\quad\textsf{this is the solution.}


I hope this helps. =)


Tags:  <em>rational equation fraction solution algebra</em>

7 0
2 years ago
Question regarding logarithms.
Eddi Din [679]

5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}

\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}

\dfrac{5^2\cdot5^2}{7^3\cdot7}=\left(\dfrac{5}{7}\right)^{x-2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{5^4}{7^4}=\left(\dfrac{5}{7}\right)^{x-2}\\\\\left(\dfrac{5}{7}\right)^4=\left(\dfrac{5}{7}\right)^{x-2}\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}

6 0
3 years ago
Read 2 more answers
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