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neonofarm [45]
3 years ago
15

Type the correct answer in each box. If necessary, use / for the fraction bar(s).

Mathematics
2 answers:
Solnce55 [7]3 years ago
8 0

Answers:

a*b = 1/2

a/b = 8/9

=============================================================

Explanation:

The horizontal bar over the 6 means the 6 goes on forever

So we have a = 0.66666... which converts to the fraction a = 2/3

The b = 0.75 converts to b = 3/4

-------------------

We can then say

a*b = (2/3)*(3/4)

a*b = (2*3)/(3*4)

a*b = 6/12

a*b = 1/2

and

a/b = (2/3) divide by (3/4)

a/b = (2/3)*(4/3)

a/b = (2*4)/(3*3)

a/b = 8/9

Note how I flipped the second fraction when I changed from a division to a multiplication.

Natalija [7]3 years ago
7 0

Answer:

this is my best guess

3.75 and 0.5

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You must first set up an equation for this problem.  Your equation is 8+3x=128.  If we subtract 8 from each side we get 3x=120.  If we then divide both sides by 3 we get x=40.  Therefore your number is 40 because 8+3*40=128.
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What is 0.5, 3/16, 0.75, 5/48 to Least to greatest?
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lena finished her assignment in 1/2 hours. then she completed her other assignment in 1/5 hours. what was the total time lena sp
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7/10 hours

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so multiply 1/2 by 5/5 to obtain 5/10

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A frog is climbing out of a well that is 12-feet deep. The frog can climb 3 feet per hour but then it rests for an hour, during
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Please please please help and solve this with steps, much help needed thank you :) 20 points for this!
leonid [27]

Answer:

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

(Please vote me Brainliest if this helped!)

Step-by-step explanation:

\frac{dy}{dx}y=x^3+2x-5x+3

\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}

  • \mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)

1. \mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:

y'\:y=x^3+2x-5x+3

2. \mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}

yy'\:=x^3-3x+3

  • N\left(y\right)\cdot y'\:=M\left(x\right)
  • N\left(y\right)=y,\:\quad M\left(x\right)=x^3-3x+3

3. \mathrm{Solve\:}\:yy'\:=x^3-3x+3:\quad \frac{y^2}{2}=\frac{x^4}{4}-\frac{3x^2}{2}+3x+c_1

4. \mathrm{Isolate}\:y:\quad y=\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}}

5. \mathrm{Simplify}

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

7 0
3 years ago
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