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3 years ago

How to solve this problem 9-0.2(4-7)^3

Mathematics

1 answer:

kkurt [141]3 years ago

7 0

Answer:

3.6

Step-by-step explanation:

9-0.2(4-7)^3

9-0.2(-3)^3

9 -0.2 (-27)

9-5.4=3.6

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3 years ago

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schepotkina [342]

Answer:

4^8

Step-by-step explanation:

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

3 0

3 years ago

List all the 3-digit numbers that fit these clues .the hundreds digit is greater than7 .the tens is 1 more than the hundreds dig

djverab [1.8K]

In this question, there are 3 conditions that need to be met
1. The number is 3-digit
2. Hundreds digit >7 (number that will fulfill it would be 8 or 9)
3. The tens is 1 more than hundred( since the hundred possibilities is 8 or 9, then 9 in hundreds can't be used since no number higher than 9)
4. The one digit <2( that mean 0, 1)

Using the list above, you can make your possible number:
890
891

3 0

3 years ago

PLEASE HELP!!!! 25 POINTS!!!!

Andrei [34K]

Answer:

0.9562

Step-by-step explanation:

Binomial probability is mathematically expressed as:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}$

Given that p=0.18, n=5 , is calculated as:

$P(X\leq 2)=P(0)+P(1)+P(2)\\\\={5\choose 0}0.18^0(1-0.18)^5+{5\choose 1}0.18^1(1-0.18)^4+{5\choose 2}0.18^2(1-0.18)^3\\\\=0.3707+0.4069+0.1786\\\\=0.9562$

Hence, the probability of no more than 2 successes in 5 trials is 0.9562

4 0

3 years ago