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3 years ago

I am a fraction that is equal to 1/6 what am i?

Mathematics

2 answers:

lyudmila [28]3 years ago

5 0

One answer could be 2/12 and another could be 3/18.

jonny [76]3 years ago

3 0

We can multiply both the numerator and denominator by ANY number (as long as you multiply both the numerator and denominator by the SAME number) to get an equivalent fraction

examples would be 2/12 (numerator and denominator multiplied by 2)3/18 (by 3)
4/24 (by 4) etc.

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Solve for x <br> -4(5 - x) = (x + 7)

garik1379 [7]

Answer:

x=9

Step-by-step explanation:

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3 years ago

Two bags of grass seed cover 1,500 square feet of ground. If Mitch’s front yard is 1,850 square feet and his back yard is 5,635

Kamila [148]

Answer:

Step-by-step explanation:

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2 years ago

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What is the best word to describe the probability that the spinner will stop on 1? justify your reasoning.

viktelen [127]

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3 years ago

at the beginning of the school year Anderson Middle School had 480 students at the end of the year it had 504 students what is t

NikAS [45]

Answer:

The number of enrollments increased by 5%.

Step-by-step explanation:

Given that:

Number of students at the beginning of the year = 480

Number of students at the end of the year = 504

Change = 504 - 480 = 24

This indicates increase.

Percent of change = $\frac{Change}{Old\ number\ of\ students}*100$

Percent of change = $\frac{24}{480}*100$

Percent of change = 5%

Hence,

The number of enrollments increased by 5%.

4 0

3 years ago

Consider f and c below. f(x, y) = x2 i + y2 j c is the arc of the parabola y = 2x2 from (−1, 2) to (1, 2) (a) find a function f

Korvikt [17]

If there is some scalar function $f(x,y)$ such that

$\nabla f(x,y)=\mathbf f(x,y)=x^2\,\mathbf i+y^2\,\mathbf j$

then we want to find $f$ such that

$\dfrac{\partial f}{\partial x}=x^2\implies f(x,y)=\dfrac{x^3}3+g(y)$
$\dfrac{\partial f}{\partial y}=y^2=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\dfrac{y^3}3+C$
$\implies f(x,y)=\dfrac{x^3}3+\dfrac{y^3}3+C$

So the vector field $\mathbf f(x,y)$ is conservative, which means the fundamental theorem applies; the line integral of $\mathbf f$ along any path $\mathcal C$ parameterized by some vector-valued function $\mathbf r(t)$ over $a\le t\le b$ is given by

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=a}^{t=b}\mathbf f(\mathbf r(t))\cdot\dfrac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt=f(\mathbf r(b))-f(\mathbf r(a))$

In this case,

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(1,2)-f(-1,2)=\dfrac23$

5 0

3 years ago