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3 years ago

Which number line shows the solution to the inequality? y - 2 < -5

Mathematics

2 answers:

xxMikexx [17]3 years ago

6 0

Answer: is the first option .

Step-by-step explanation:

Tatiana [17]3 years ago

5 0

For this case we have the following inequality:
$y - 2 \ \textless \ -5
$
We will solve the inequality algebraically.
We have then:
We add 2 to both sides of the inequality:
$y - 2 + 2 \ \textless \ -5 + 2
$
Rewriting we have:
$y \ \textless \ - 3
$
Therefore the solution is:
(-inf, -3)
Answer:
(-inf, -3)
See attached image.

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8=2 missing exponent

Verizon [17]

The exponent is 3. 2 to the 3rd power is 8

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3 years ago

g What is the probability of being dealt exactly three of a kind (like three kings or three 7’s, etc.) in a five card hand from

Elena L [17]

Answer: 0.02257

Step-by-step explanation:

Given : Total cards in a deck = 52

Number of ways to select any 5 cards : $^{52}C_5$

Since , there are total 13 kinds of card (includes Numbers from 2 to 9 and Ace , king, queen and jack).

Of each kind , there are 4 cards.

Number of ways to select three cards in a five card hand of a single kind : $^{4}C_3\times^{48}C_2$

Number of ways to select three cards in a five card hand of a exactly three of a kind : $13\times^{4}C_3\times^{48}C_2$

Now , the required probability = $\dfrac{13\times^{4}C_3\times^{48}C_2}{^{52}C_5}$

$=\dfrac{13\times4\times\dfrac{48!}{2!46!}}{\dfrac{52!}{5!47!}}\\\\\\=\dfrac{58656}{2598960}$

$=0.022569027611\approx0.02257$

∴ The probability of being dealt exactly three of a kind (like three kings or three 7’s, etc.) in a five card hand from a deck of 52 cards= 0.02257

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3 years ago

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

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2 years ago

A line passes through the points (1.4) and (2, 2). What is the equation of this line?

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