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bija089 [108]
4 years ago
12

Factor completely x^2-48

Mathematics
1 answer:
fredd [130]4 years ago
5 0

Answer: (x-7)(x+7)


Step-by-step explanation:

To solve this problem you must apply the proccedure shown below;

- Make the equation equal to zero:

x^{2}-49=0

-Add 49 at both sides:

x^{2}=49

- Now, you must take the square root of both sides of the equation, then you obtain the following roots:

x=\sqrt{49}\\x_1=7\\x_2=-7

- Then you have:

x^{2}-49=(x-7)(x+7)

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Please help!!!!!! 5 points please
Alex73 [517]
\left \{ {{-x + 2y = 6} \atop {6y=x + 18}} \right.

First, we should rearrange the first equation to set it equal to y:

-x + 2y = 6
2y = 6 + x
y = 3 + 0.5x

Now, we can plug this into the second equation for y and solve for x:

6y = x + 18
6(3 + 0.5x) = x + 18
18 + 3x = x + 18
3x = x
2x = 0
x = 0

The x-value of the solution is 0.
6 0
3 years ago
PLEASE HELP!!!!!! The line of symmetry for the quadratic equation y = ax 2 - 8x - 3 is x = 2. What is the value of "a"?
nataly862011 [7]
y= ax^{2} -8x-3

1.

the line of symmetry is x=2, means that the x coordinate of the vertex is x=2.

the point x=2 is the midpoint of the roots x_1 and x_2. 

so 
\frac{x_1+x_2}{2}=2

x_1+x_2=4

Remark: in the x-axis, if c is the midpoint of a and b, then c= \frac{a+b}{2}


2.
since x_1 and x_2 are roots 

a(x_1)^{2} -8(x_1)-3=0 and a(x_2)^{2} -8(x_2)-3=0

3.
equalizing:

a(x_1)^{2} -8(x_1)-3=a(x_2)^{2} -8(x_2)-3

a(x_1)^{2} -8(x_1)=a(x_2)^{2} -8(x_2)

a(x_1)^{2}-a(x_2)^{2} =8(x_1) -8(x_2)

in the left side factorize a, in the left side factorize 8:

a[(x_1)^{2}-(x_2)^{2}] =8(x_1 -x_2)

in the right side use the difference of squares formula:

a(x_1 -x_2)(x_1 +x_2) =8(x_1 -x_2)

simplify by (x_1 -x_2)

a(x_1 +x_2) =8

substitute (x_1 +x_2) with 4:

a*4 =8

a=2


Answer: C)2

5 0
4 years ago
What is the geometric mean of 10 and 40
Varvara68 [4.7K]

the geometric mean is 20

the geometric mean of 2 numbers a and b is \sqrt{ab}

\sqrt{10(40)} = \sqrt{400} = 20


8 0
3 years ago
You buy a used car for $10,000. The value of the car decreases by22% each year. Approximately how much is the car worth in 3 yea
IRINA_888 [86]

Answer:

The value will be $3,400 after 3 years

Step-by-step explanation:

Convert the percentage into the cash decrease each year, multiply that by 3 and then subtract $10,000 by the sum.

.22 x 10,000 = 2,200

2,200 x 3 = 6,600

10,000 - 6,600 = 3,400

6 0
3 years ago
Proof by induction on the number of horses: Basis Step. There is only one horse. Then clearly all horses have the same color. In
Novosadov [1.4K]

Answer:

Claiming mathematical induction, of the statement: "all horses are the same color", the theorem is a counterfeit paradox sustained by mistaken  demonstrations.

Step-by-step explanation:

”that is a horse of a different  color” was a familiar expression in the middle of the last century, meaning that something is quite different from normal or common expectation, but George Polya, a great mathematician provided proof that there is no horse of a different color:

Theorem: "All horses are the same color"

Proof (by induction on the number of horses):

- Base Case: P(1) is undoubtedly true, as having only one horse, then all horses have the same color.

- Inductive Hypothesis: Assume P(n), which is the statement that n horses all have the same color.

- Inductive Step: Given a set of n+1 horses {h1,h2,...,hn+1}, we can eliminate the last horse in the serie  and use the inductive hypothesis onlky to the first n horses {h1,...,hn}, deducing that they all have  the same color. The same way, the conclusion may be that the last n horses {h2,...,hn+1} all have the same  color. But the “middle” horses {h2,...,hn} (i.e., all but the first and the last) belong to both of  these series, so they have the same color as horse h1 and horse hn+1. It follows, therefore, that all n+1  horses have the same color. Therefore, using the principle of induction, all horses have the same color.

It is clear that, it is not true that all horses are of the same color, so where is the mistake in our induction  proof? It is tempting to blame the induction hypothesis. But even though the induction hypothesis is false  (for n ≥ 2), that is not the mistaken reasoning. The real flaw in the proof is that the induction step is valid for a “typical”  value of n, say, n = 3. The flaw, however, is in the induction step when n = 1. In this case, for n+1 = 2  horses, there are no “middle” horses, this makes the argument to collapse.

7 0
3 years ago
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