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ollegr [7]
2 years ago
9

Solve quadratic equation 10x^2+x-2=0

Mathematics
1 answer:
zhuklara [117]2 years ago
8 0

Answer:

\{-\frac{1}{2} \} \cup \{\frac{2}{5}\}

Step-by-step explanation:

10x^2 + x - 2 = 0; \\ a = 10, b = 1, c = -2; \\ D = b^2 - 4ac = 1^2 - 4 * 10 * (-2) = 1 + 80 = 81 = 9^2, > 0; \\ x_{1, 2} = \frac{-b \pm \sqrt{D}}{2a} = \frac{-1 \pm \sqrt{9^2}}{2 * 10} = \frac{-1 \pm 9}{20} = \left \ [ {{\frac{-1 - 9}{20} = -\frac{10}{20} = -\frac{1}{2}  } \atop {\frac{-1 + 9}{20} = \frac{8}{20} = \frac{2}{5}}} \right.

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3 years ago
Five times my age 4 years ago is the same as 3 times my age in 2 years." How old is Elmo now?
zheka24 [161]

Answer:

elmo is 3 right now

Step-by-step explanation:

5x -4 = 3x+2

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7 0
3 years ago
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When a number is decreased by 20% of itself, the result is 128. what is the number?
andriy [413]
The number was 160 because 5/4•128=160. 5/4 represents inverting 80% and multiplying it by the number to determine how the result became 128. For checking over all you have to do is do 4/5 or 80%•160 which shall give you 128.
4 0
3 years ago
Point F is on circle C. What is the length of line segment GF?
Levart [38]
The answer:
the full question is as follow:
<span>Point F is on circle C. What is the length of line segment GF?

12.5  units
15.0  units
17.5  units
20.0  units

according to the image, GF= GC + CF

CF is the radius of the circle, so it is CF =CE = 7.5
all that we want to find is the value of GC
let's consider the triangle GEC. This is a right triangle, so for finding GC, we can apply Pythagorean theorem

that is, GE² + EC² = GC², and from this, we have GC = sqrt(</span>GE² + EC² )

GC = sqrt(10² + 7.5² )=sqrt(56.25)=12.5
<span>
therefore, </span>GF= GC + CF=12.5 + 7.5 = 20.0
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8 0
3 years ago
Read 2 more answers
Evaluate C_n.xP^xQn-x For the given n=7, x=2, p=1/2
r-ruslan [8.4K]

Answer:

The value of given expression is \frac{21}{128}.

Step-by-step explanation:

Given information: n=7, x=2, p=1/2

q=1-p=1-\frac{1}{2}=\frac{1}{2}

The given expression is

C(n,x)p^xq^{n-x}

It can be written as

^nC_xp^xq^{n-x}

Substitute n=7, x=2, p=1/2 and q=1/2 in the above formula.

^7C_2(\frac{1}{2})^2(\frac{1}{2})^{7-2}

\frac{7!}{2!(7-2)!}(\frac{1}{2})^2(\frac{1}{2})^{5}

\frac{7!}{2!5!}(\frac{1}{2})^{2+5}

\frac{7\times 6\times 5!}{2\times 5!}(\frac{1}{2})^{2+5}

21(\frac{1}{2})^{7}

\frac{21}{128}

Therefore the value of given expression is \frac{21}{128}.

7 0
3 years ago
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