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2 years ago

6

Bill is filling a box up with bottle caps. The box is 1 1/2 feet long, 15 inches wide, and 9 inches high. What is the volume of

the box in cubic inches

1 answer:

Doss [256]2 years ago

5 0

11/2 so 15 inches high cubic box will get more high I guess

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A rectangular photograph is nine centimeters wide and fourteen centimeters long. The photograph is enlarged by increasing the le

koban [17]

Answer:

Width=13cm

Length=18cm

Step-by-step explanation:

Width of the Initial Photograph=9cm

Length of the Initial Photograph=14cm

If width and length are increased by an equal amount, say x

New Width=9+x

New Length=14+x

Area of the new photo is 108 square centimeters greater than the area of the original photo.

Area of Original Photo=14*9=126cm²

Aeea of the New Photo=126+108=234cm²

Therefore:

(9+x)(14+x)=234

126+9x+14x+x²-234=0

x²+23x-108=0

x²+27x-4x-108=0

x(x+27)-4(x+27)=0

(x+27)(x-4)=0

x+27=0 or x-4=0

x=-27 or x=4

Since x≠-27

x=4cm

The dimensions of the new photo are:

Width=9+x=9+4=13cm

Length=14+x=14+4=18cm

4 0

3 years ago

What is the prime factorization ?? What is the number and explain ??

dimaraw [331]

The number is already in prime factored form. The prime factors are 2, 3, 5, and 7. If we multiply everything out then we end up with 2100

6 0

3 years ago

Help pls no links I'll give briliantest to who ever gives me the right answer

gladu [14]

Answer:

A. 57 7/9

Step-by-step explanation:

4 1/3 x 3 1/3 = 14 4/9

14 4/9 x 4 = 57 7/9

6 0

2 years ago

How do I put these equations into vertex form?

kenny6666 [7]

I found my notes for this exact paper and I have that first problem written. hopefully this helps

4 0

2 years ago

For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

1.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

2.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

3.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

4.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x^2\implies f(x,y,z)=-x^3+g(y,z)$

$\dfrac{\partial f}{\partial y}=5y^2=\dfrac{\partial g}{\partial y}\implies g(y,z)=\dfrac53y^3+h(z)$

$\dfrac{\partial f}{\partial z}=5z^2=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac53z^3+C$

$\implies\boxed{f(x,y,z)=-x^3+\dfrac53y^3+\dfrac53z^3+C}$

so $\vec F$ is conservative.

4 0

3 years ago