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2 years ago

On Tuesday, 0.48 of the students at West Middle School rode the bus to school. What fraction of the students

Mathematics

1 answer:

kotykmax [81]2 years ago

4 0

Answer:

ygyggygygygygghggdhgfegfgrgfhgrfrfvjegfhrggfhgrj

Step-by-step explanation:

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7.) Find the Unit Rate: 32 flowers per 2 vases (flowers per vase)

IgorLugansk [536]

Answer:

Step-by-step explanation:

I JUST TOOK THE TEST YW

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2 years ago

Please Simplify 60/105 to lowest form

svlad2 [7]

4/7. Divide the numerator and denominator by 15.

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3 years ago

What is the image of the point (-6,-8)(−6,−8) after a rotation of 90^{\circ}90

mrs_skeptik [129]

Answer:

(8,-6)

Step-by-step explanation:

The rule is (x,y) becomes (-y,x). You can find all of these rules online.

7 0

1 year ago

Can someone please help me

uranmaximum [27]

If we have y = f(x), then vertically stretching that by a factor of k gives y = k*f(x) as the new function. This means we simply stick a 3 in front of the 2^x to get our answer.

Answer: Choice C) g(x) = 3*2^x

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3 years ago

A cardboard box without a lid is to have a volume of 5,324 cm3. Find the dimensions that minimize the amount of cardboard used.

romanna [79]

The area of the box is

$A(x,y,z)=xy+2xz+2yz$

which we want to minimize subject to the constraint $xyz=5324$ .

The Lagrangian is

$L(x,y,z)=xy+2xz+2yz+\lambda(xyz-5324)$

with critical points where the partial derivatives are 0:

$L_x=y+2z+\lambda yz=0$

$L_y=x+2z+\lambda xz=0$

$L_z=2x+2y+\lambda xy=0$

$L_\lambda=xyz-5324=0$

Notice that

$L_y-L_x=(x-y)+\lambda(xz-yz)=0\implies(x-y)(1+\lambda z)=0\implies x=y\text{ or }z=-\dfrac1\lambda$

Substituting the latter into either $L_x=0$ or $L_y=0$ will end up suggesting that $\lambda$ is infinite, so we throw out this case.

If $x=y$ , then

$L_z=0\implies4x+\lambda x^2=0\implies x=0\text{ or }x=-\dfrac4\lambda$

We ignore the case where $x=0$ because that would make the volume 0. Then

$x=y=-\dfrac4\lambda\text{ and }L_x=0\implies-\dfrac4\lambda+2(1331\lambda^2)+\lambda\left(-\dfrac4\lambda\right)(1331\lambda^2)=0$

$\implies2662\lambda^3+4=0$

$\implies\lambda=-\dfrac{\sqrt[3]{2}}{11}$

so we have one critical point at

$(x,y,z)=\left(22\sqrt[3]{4},22\sqrt[3]{4},\dfrac{11}{2\sqrt[3]{2}}\right)\approx(34.9228,34.9228,4.3654)$

which give a minimum area of about 1829.41 sq. cm.

6 0

2 years ago