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3 years ago

Jon red a rides in his pocket when he found two pennies

Mathematics

1 answer:

Sliva [168]3 years ago

8 0

Answer:

Step-by-step explanation:

2 pennies equal 2 cents.

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Solve system of equation using elimination by addition.<br><br> Will give brainliest!!

DedPeter [7]

Answer:

$x = 2, y = -3$

Step-by-step explanation:

Add the 2 equations together, then solve for x.

$(2x + 2y) + (3x - 2y) = (-2) + (12)\\ 2x + 3x + 2y - 2y = -2 + 12\\5x = 10\\x = 2$

Now that we know the value of x, we can easily find y by substitution.

$3x - 2y = 12\\3(2) - 2y = 12\\6 - 2y = 12\\-2y = 6\\y = -3$

These are the answers

- Kan Academy Advance

5 0

2 years ago

Caleb's monthly bank statement showed the following deposits and withdrawals:

Crazy boy [7]

Answer:

$174.72

Step-by-step explanation:

105.88+43.50+36.61+92.78-38.19-65.86=174.72

6 0

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

For , find the value of x for which .

Mice21 [21]

Uh what did you mean to upload a picture or somthing

8 0

3 years ago

How could I write the fractions 8/100 3/5 and 7/10 from least to greatest?

iVinArrow [24]

Get them to have a common denominator. 100 is the common denominator, so change 7/10 and 3/5 to have a denominator of 100. 7/10=70/100 and 3/5=60/100. So from least to greatest it will go 8/100, 3/5, 7/10.

5 0

3 years ago

Read 2 more answers