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yan [13]
2 years ago
7

What method would you choose to solve the equation 2x2 – 7 = 9? Explain why you chose this method.

Mathematics
2 answers:
arlik [135]2 years ago
8 0

Sample Response/Explanation: I would use the square root property of equality. Because there is no x term, I would just add and divide to get x2 by itself. Then I would take the square root.

oksano4ka [1.4K]2 years ago
7 0

I would add 7 to get ...

... 2x² = 16

then divide by 2 to get

... x² = 8

Now, I would take the square root, recognizing that both positive and negative values will solve the problem.

... x = ±√8

The root can be simplified, so we have ...

... x = ±2√2

_____

The above method seemed the most obvious method to me. One could use the quadratic formula, but that seems to involve more operations.

... 2x² -16 = 0 . . . . subtract 9

... x = (-0 ± √(0² -4·2·(-16)))/(2·2) . . . . put the coefficients into the formula

... x = ±(√128)/4 = ±√8 = ±2√2 . . . . . simplify the result

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7. The probability that a person living in a certain city owns a dog is estimated to be 0.3. Find the probability that the tenth
STALIN [3.7K]

Answer:lol can somebody please answer this I need an answer I have this question

Step-by-step explanation:

4 0
2 years ago
ABC is a triangle and PQ is parallel to BC.<br>BC=12.6cm, PQ=8.4 cm and AQ= 7.2 cm.<br>Find AC​
nlexa [21]

With a line parallel to a side through a triangle we get similar triangles so a proportion.

AC : BC = AQ : PQ

AC = BC AQ / PQ

AC = 12.6 (7.2) / 8.4

AC = 10.8 cm

Answer: 10.8 cm

6 0
2 years ago
In hypothesis​ testing, does choosing between the critical value method or the​ P-value method affect your​ conclusion? Explain.
Artist 52 [7]

Answer:​

The correct answer is No.

Choosing between the critical value method or the P-value method does not affect one's conclusion because both methods look at the probability of the test​ statistic's and its level of significance .

Given the methodology utilized by both methods, they usually arrive at the same conclusion.

Cheers!

8 0
2 years ago
Circle J is located in the first quadrant with center (a, b) and radius s. Felipe transforms Circle J to prove that it is simila
iVinArrow [24]
The correct answer for the question that is being presented above is this one: "c. translate circle J by (x-a, y-b) and dilate by a factor of t/s." Circle J is located in the first quadrant with center (a, b) and radius s. Felipe transforms Circle J to prove that it is similar to any circle centered <span>at the origin with radius r.</span>
3 0
2 years ago
If p is a polynomial show that lim x→ap(x)=p(a
Lostsunrise [7]

Let p(x) be a polynomial, and suppose that a is any real number. Prove that

lim x→a p(x) = p(a) .

 

Solution. Notice that

 

2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .

 

So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x – 2.

 

Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| < 1, so −2 < x < 0. In particular |x| < 2. So

 

|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|

= 2|x|^3 + 5|x|^2 + |x| + 2

< 2(2)^3 + 5(2)^2 + (2) + 2

= 40

 

Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2 + x − 2| < ε/40 · 40 = ε.

3 0
2 years ago
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