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2 years ago

5

P Josh wants to determine the number of loaves of bread he can bake using the 13 and one-third cups of flour he has. If each loa

f requires 3 and four-fifths cups of flour, how many loaves can Josh bake?
3 loaves of bread
4 loaves of bread
50 loaves of bread
51 loaves of bread

1 answer:

ruslelena [56]2 years ago

4 0

Answer:

the 13 and one-third cups of flour he has. If each loaf requires 3 and four-fifths cups of flour, how many loaves can Josh bake?

3 loaves of bread

4 loaves of bread

50 loaves of bread

51 loaves of bread

Step-by-step explanation:

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Volume of cube=side³
ok, so you need to know the difference or sum of cubes
a³+b³=(a+b)(x²-xy+y²)
so

(4p)³+(2q²)³=
(4p+2q²)((4p)²-(4p)(2q²)+(2q²)²)=
(4p+2q²)(16p²-8pq²+4q⁴)

3rd option

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2 years ago

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

Bas_tet [7]

Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

$\bar{x}=\frac{21}{168}*40=5$

Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

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blondinia [14]

Based on the given figure above, we can conclude that the triangle is an isosceles triangle. By definition, an isosceles triangle is a triangle that has at least two equal sides. Since this is an isosceles triangle, 8x-10 =6x. Now we can solve for x. So,
8x-10 =6x
8x-6x = 10
2x =10
x= 5.
Therefore, the value of x in the figure is 5. Hope this is the answer that you are looking for.

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