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Murrr4er [49]
3 years ago
14

300 ml over 15 min is what rate of ml per hour

Mathematics
1 answer:
Mama L [17]3 years ago
3 0

Answer:

1200 ml per hour

Step-by-step explanation:

To compute what rate is 300 ml over 15 mins as a rate of ml per hour, we do a rule of 3, using a variable x as that amount of ml we don't know yet. We should have everything in the same units, so instead of writing 1 hour we write 60 minutes:

\frac{300~ml}{15~min}=\frac{x~ml}{60~mins}

Now we solve for x:

\frac{300~ml}{15~min}\cdot 60~mins=x~ml

\frac{18000~ml}{15} =x~ml

1200~ml =x~ml

And so, now that we know the value of x, the rate we wanted to find is

\frac{1200~ml}{60~mins}

Which is just 1200 ml per hour.

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Rewrite the quadratic function in vertex form.<br> Y=2x^2+4x-1
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Answer:

\large\boxed{y=2(x+1)^2-3}

Step-by-step explanation:

The vertex form of an equation of a parabola:

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

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We have

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

We must use the formula: (a+b)^2=a^2+2ab+b^2\qquad(*)

2\left(x^2+2(x)(1)-\dfrac{1}{2}\right)=2\bigg(\underbrace{x^2+2(x)(1)+1^2}_{(*)}-1^2-\dfrac{1}{2}\bigg)\\\\=2\left((x+1)^2-1-\dfrac{1}{2}\right)=2\left((x+1)^2-\dfrac{3}{2}\right)

Use the distributive formula a(b + c) = ab + ac

2(x+1)^2+2\left(-\dfrac{3}{2}\right)=2(x+1)^2-3

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3 years ago
What is the midpoint of the segment shown below
andreev551 [17]

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5,6 A

Step-by-step explanation:

6 0
3 years ago
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What is the slope of the line whose equation is y−4=5/2(x−2)?
postnew [5]

Answer:

m=\dfrac{5}{2}

Step-by-step explanation:

If the equation of the line is

y=mx+b,

then m represents the slope of the line and b represents the y-intercept of the line. This equation is called the equation of the line in the slope form.

Rewrite the equation of the line in the slope form

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

Thus, the slope of the line is

m=\dfrac{5}{2}

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A sample size of n=12 is a simple random sample selected from a normally distributed population. Find the critical value t* corr
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Answer:

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Step-by-step explanation:

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Find all solutions of the equation: 2cos^2x-cosx=1
Art [367]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3166243

——————————

Solve the trigonometric equation:

     \mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}


Make a substitution:

     \mathsf{cos\,x=t\qquad (-1\le t\le 1)}

and the equation becomes

     \mathsf{2t^2-t-1=0}


Rewrite conveniently  – t  as  + t – 2t,  and then factor the left-hand side by grouping:

      \mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}


Factor out  2t + 1:

     \mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}


Substitute back for  t = cos x:

     \begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}


Therefore,

     \begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}

where  k  is an integer.


Solution set:   

\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}


I hope this helps. =)

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