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3 years ago

7

mya has ballet lessons every sixth day and swimming every fourth day. today she has both lessons. in how many days will mya have

both lessons on the same day again?

1 answer:

ra1l [238]3 years ago

5 0

To determine when Mya will have both lessons again on the same day, you will list the multiples of each number of days because to show every 4 or 6 days, you will count by 4's and 6's.

When you get to the first number that is the same, that will be the next time she will have both lessons again. This is called the least common multiple (LCM).

4, 8, 12, 16, 20, ...
6, 12, 18, 24

In 12 days she will have both lessons again.

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Suppose the students construct a 95% confidence interval for the true weight of their rocks. Whose interval do you expect to be

MakcuM [25]

Answer:

(i) The student who weighed the rock 20 times.

Step-by-step explanation:

The margin of error of a confidence interval is:

$M = z*\frac{\sigma}{\sqrt{n}}$

In which z is related to the confidence level, $\sigma$ is the standard deviation of the population and n is the size of the sample.

From the equation above, the larger the size of the sample, the lower the margin is, that is, the narrower the interval is.

In this question:

A student weighed the rock 20 times and other the 5. From above, the one who weighed the rock 20 times will have the more precise interval.

So the correct answer is:

(i) The student who weighed the rock 20 times.

7 0

3 years ago

a parabola has x intercepts of -1 and 3 it passes through the point (1,6) write the equation if the parabola

matrenka [14]

$\boxed{f(x)=-\frac{3}{2}x^2+3x+\frac{9}{2}}$

<h2>Explanation:</h2>

The Standard Form of the equation of a parabola is given by the form:

$y=ax^2+bx+c$

As y is a function of x, then we can write:

$y=f(x)=ax^2+bx+c \\ \\ a, \ b, c \ Real \ Coefficients$

Also, we can write $f(x)$ in factored form as:

$f(x)=a(x-x_{1})(x-(x_{2}) \\ \\ x_{1} \ and \ x_{2} \ Are \ the \ roots$

In this case:

$x_{1}=-1 \\ \\ x_{2}=3$

So:

$f(x)=a(x-(-1))(x-3) \\ \\ f(x)=a(x+1)(x-3)$

We also know that the parabola passes through the point (1,6):

$f(1)=6=a(1+1)(1-3) \\ \\ Isolating \ a: \\ \\ 6=a(2)(-2) \\ \\ 6=a(-4) \\ \\ a=-\frac{6}{4} \\ \\ a=-\frac{3}{2}$

Finally, the equation of the parabola is:

$f(x)=-\frac{3}{2}(x+1)(x-3)$

Or applying Distributive Property:

$f(x)=-\frac{3}{2}\left(x^2-2x-3\right) \\ \\ \boxed{f(x)=-\frac{3}{2}x^2+3x+\frac{9}{2}}$

<h2>Learn more:</h2>

Piecewise-defined function: brainly.com/question/12169258

#LearnWithBrainly

3 0

3 years ago

Which of the following fractions has the greatest value?

strojnjashka [21]

Answer:

5/8

Step-by-step explanation:

5/8 has the greatest value

4 0

3 years ago

Read 2 more answers

Simplify the following:

Charra [1.4K]

1) 4x + 5
2) 14w2 - 4w
3) -x + 19
4) 19x - 16
5) -x2 - x
6) 11a2b - 12ab2

5 0

3 years ago

PLs help 50 PTS!!!!! PLEASE ILL GIVE BRAINLIEST!!!!!

Nookie1986 [14]

Answer:

$\large\boxed{y=\dfrac{1}{4}x^2-x-4}$

Step-by-step explanation:

The equation of a parabola in vertex form:

$y=a(x-h)^2+k$

<em>(h, k)</em><em> - vertex</em>

The focus is

$\left(h,\ k+\dfrac{1}{4a}\right)$

We have the vertex (2, -5) and the focus (2, -4).

Calculate the value of <em>a</em> using $k+\dfrac{1}{4a}$

<em>k = -5</em>

$-5+\dfrac{1}{4a}=-4$ <em>add 5 to both sides</em>

$\dfrac{1}{4a}=1$ <em>multiply both sides by 4</em>

$4\!\!\!\!\diagup^1\cdot\dfrac{1}{4\!\!\!\!\diagup_1a}=4$

$\dfrac{1}{a}=4\to a=\dfrac{1}{4}$

Substitute

$a=\dfrac{1}{4},\ h=2,\ k=-5$

to the vertex form of an equation of a parabola:

$y=\dfrac{1}{4}(x-2)^2-5$

The standard form:

$y=ax^2+bx+c$

Convert using

$(a-b)^2=a^2-2ab+b^2$

$y=\dfrac{1}{4}(x^2-2(x)(2)+2^2)-5\\\\y=\dfrac{1}{4}(x^2-4x+4)-5$

<em>use the distributive property: a(b+c)=ab+ac</em>

$y=\left(\dfrac{1}{4}\right)(x^2)+\left(\dfrac{1}{4}\right)(-4x)+\left(\dfrac{1}{4}\right)(4)-5\\\\y=\dfrac{1}{4}x^2-x+1-5\\\\y=\dfrac{1}{4}x^2-x-4$

3 0

3 years ago