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3 years ago

26. 50 students sit in a class room. Each student

1 answer:

8 0

<h2><em>each student requires 9m² of floor </em></h2><h2><em>and given the no. of students is 50 </em></h2><h2><em>So the total area of the room is 9m²X50=450m² </em></h2><h2><em>Given the length of the room is 25m </em></h2><h2><em>so the Breadth is =450/25=18m </em></h2><h2><em> </em></h2><h2><em>each student requires 108m³ of space </em></h2><h2><em>so total volume of the room is 108X50=5400m³ </em></h2><h2><em>we know that : Volume=Area X h </em></h2><h2><em> so⇒5400=450Xh </em></h2><h2><em> ⇒h=5400/450= 12m</em></h2><h2><em /></h2><h2><em>HOPE IT HELPS (◕‿◕✿)</em></h2>

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In △ABC, a=31, b=22, and c=18. Identify m∠A rounded to the nearest degree. HELP ME PLEASE!!

4vir4ik [10]

Answer:

The measure of angle A is $101\°$

Step-by-step explanation:

we know that

Applying the law of cosines

$a^{2} =b^{2}+c^{2}-2(b)(c)cos(A)$

substitute the values and solve for cos(A)

$31^{2} =22^{2}+18^{2}-2(22)(18)cos(A)$

$cos(A)=[22^{2}+18^{2}-31^{2}]/(2(22)(18))\\ \\cos(A)=-0.193182\\ \\A=arccos(-0.193182)=101\°$

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4 years ago

Find the smallest value of k such that 16k is a perfect cube.

IrinaK [193]

Answer:

Step-by-step explanation:

The smallest perfect cube would be 0 since it cannot be negative and zero raised to the third power would still be 0

3 0

3 years ago

Read 2 more answers

Consider the function V=g(x), where g(x) =x(6-2x)(8-2x), with x being the length of a cutout in cm and V being the volume of an

Andrej [43]

Answer:

The maximum volume of the open box is 24.26 cm³

Step-by-step explanation:

The volume of the box is given as $V=g(x)$ , where $g(x)=x(6-2x)(8-2x)$ and $0\le x\le3$ .

Expand the function to obtain:

$g(x)=4x^3-28x^2+48x$

Differentiate wrt x to obtain:

$g'(x)=12x^2-56x+48$

To find the point where the maximum value occurs, we solve

$g'(x)=0$

$\implies 12x^2-56x+48=0$

$\implies x=1.13,x=3.54$

Discard x=3.54 because it is not within the given domain.

Apply the second derivative test to confirm the maximum critical point.

$g''(x)=24x-56$ , $g''(1.13)=24(1.13)-56=-28.88\:$

This means the maximum volume occurs at $x=1.13$ .

Substitute $x=1.13$ into $g(x)=x(6-2x)(8-2x)$ to get the maximum volume.

$g(1.13)=1.13(6-2\times1.13)(8-2\times1.13)=24.26$

The maximum volume of the open box is 24.26 cm³

See attachment for graph.

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4 years ago

What will happen to me if I get a 0 on my homework and it's worth 20% of my grade? This is the 3rd time this happened.

IrinaVladis [17]

Your overall grade will go down. I'd try to ask your teacher if you can make them up.

7 0

3 years ago

GIVING 100 POINTS!!!!

IrinaK [193]

Answer:

As per dot plots we see the distribution of prices is close but majority of prices are concentrated in different zones. So MAD would be more similar by the look.

<u>Let's verify</u>

<h3>Neighborhood 1</h3>

55, 55, 60, 60, 70, 80, 80, 80, 90, 120

(55*2+ 60*2+ 70+ 80*3 + 90+ 120)/10 = 75

(20*2+15*2+5+5*3+15+45)/10 = 15

<h3>Neighborhood 2</h3>

100, 110, 110, 110, 120, 120, 120, 140, 150, 160

(100 + 110*3+ 120*3+ 140 + 150+ 160)/10 = 124

(24+14*3+4*3+16*3+16+26+36)/10 = 20.4

As we see the means are too different (75 vs 124) than MADs (15 vs 20.4).

7 0

3 years ago

Read 2 more answers