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Stella [2.4K]
3 years ago
13

Solve the equation for x∈Z -x² +8x -14 ≥ 0

Mathematics
2 answers:
kenny6666 [7]3 years ago
6 0
ax^2+bx+c=0\\\Delta=b^2-4ac\\if\ \Delta < 0\ then\ no\ solutions\\if\ \Delta=0\ then\ one\ solution:x=\frac{-b}{2a}\\if\ \Delta > 0\ then\ two\ solutions:x=\frac{-b\pm\sqrt\Delta}{2a}\\================================\\\\-x^2+8x-14\geq0\ \ \ \ |multiply\ both\ sides\ by\ (-1)\ \{change\ \geq\ on\ \leq\}\\\\x^2-8x+14\leq0\\\\a=1;\ b=-8;\ c=14\\\\\Delta=(-8)^2-4\cdot1\cdot14=64-56=8 > 0\\\sqrt\Delta=\sqrt8=\sqrt{4\cdot2}=\sqrt4\cdot\sqrt2=2\sqrt2

x_1=\frac{-(-8)-2\sqrt2}{2\cdot1}=\frac{8-2\sqrt2}{2}=\frac{8}{2}-\frac{2\sqrt2}{2}=\boxed{4-\sqrt2}\\\\x_2=\frac{-(-8)+2\sqrt2}{2\cdot1}=\frac{8+2\sqrt2}{2}=\frac{8}{2}+\frac{2\sqrt2}{2}=\boxed{4+\sqrt2}\\\\============================

ax^2+bx+c=0\\\\if\ a > 0\ then\ the\ parabola\ o pen\ up\\if\ a < 0\ then\ the\ parabola\ o pen\ down\\========================\\\\a=1 > 0-therefore\ o pen\ up\ (look\ at\ the\ picture)\\\\===============================\\\\Answer:x\in\left


Solutions\ in\ \mathbb{Z}:\\\\\sqrt2\approx1.4\\\\threfore:4-\sqrt2\approx4-1.4=2.6\ and\ 4+\sqrt2\approx4+1.4=5.4\\\\look\ at\ the\ second\ picture:x=3\ or\ x=4\ or\ x=5\ (x\in\{3;\ 4;\ 5\})

tensa zangetsu [6.8K]3 years ago
5 0
-x^2 +8x -14 \geq 0\\&#10;-(x^2-8x+14)\geq0\\&#10;-(x^2-8x+16-2)\geq0\\&#10;-((x-4)^2-2)\geq0\\&#10;-(x-4)^2+2\geq0\\&#10;-(x-4)^2\geq-2\\&#10;(x-4)^2\leq2\\&#10;x-4 \leq \sqrt2 \wedge x-4\geq-\sqrt2\\&#10;x\leq 4+\sqrt2 \wedge x\geq4-\sqrt2\\&#10;x\in[4-\sqrt2,4+\sqrt2]\\\\&#10;x\in[4-\sqrt2,4+\sqrt2] \wedge x\in\mathbb{Z}\\&#10;\boxed{x=\{3,4,5\}}&#10;&#10;&#10;
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Please help, i don’t understand this at all.
zubka84 [21]
So, standard form basically takes the shape of Ax+By=C. You want all of your variables to be on the left side and your constant on the right. There can also be no fractions!

In your case, since you didn't mention a y value for y, your line is y=-3/2x+6

first we get rid of the fraction by multiplying both sides of the equation by 2:
(2)y= (-3/2x+6)(2) and get
2y=-3x+12
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3 years ago
The number square root of 3 goes on forever with no repeating pattern therefore it is rational?
vova2212 [387]

Answer:

False

Step-by-step explanation:

A number that can expressed as \frac{p}{q},

Where, p and q are integers such that q ≠ 0, is called a rational number.

Also, a decimal number with repeating pattern is a rational number,

( for eg : 0.333.... = \frac{3}{9} )

Since,

√3 = 1.73205080757.....

∴ There is no repeating pattern

Thus, √3 is not a rational number.

Given statement is FALSE.

8 0
3 years ago
Show by substitution whether the number r is a solution of the corresponding quadratic equation.
Stels [109]

Answer:

It's choice B.

Step-by-step explanation:

When r= -11,

x^2 + 8x - 33 =  (-11)^2 - 88 - 33

= 121 - 88 - 33

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3 0
2 years ago
A company makes rubber rafts. 12% of them develop cracks within the first month of operation. 27 new rafts are randomly sampled
rewona [7]

The probability that the number of tested rafts that develop cracks is no more than 3 is <u>.00006</u>.

The true proportion, p for the population is given to 0.12.

Thus, the mean, μ, for the sample = np = 27*0.12 = 3.24.

The sample size, n, given to us is 27.

Thus, the standard deviation, s, for the sample can be calculated using the formula, s = √{p(1 - p)}/n.

s = √{0.12(1 - 0.12)}/27 = √0.003911 = 0.0625389.

We are asked to calculate the probability that the number of tested rafts that develop cracks is no more than 3, that is, we need to calculate P(X ≤3).

P(X ≤ 3)

= P(Z ≤ {(3 - 3.24)/0.0625389) {Using the formula z = (x - μ)/s}

= P(Z ≤ -3.8376114706)

= .00006 {From table}.

Thus, the probability that the number of tested rafts that develop cracks is no more than 3 is <u>.00006</u>.

Learn more about sampling distributions at

brainly.com/question/15507495

#SPJ4

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it's B

I just took the test

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