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sergij07 [2.7K]
4 years ago
8

Mr. ishimoto ordered x new math books and y new workbooks for his class. the total weight of the box of books cannot be more tha

n 50 pounds. if each math book weighs 3.2 pounds and each workbook weighs 0.8 pounds, which inequality represents the maximum number of each type of book that can be shipped in a single box?3.2x 0.8y < 503.2x 0.8y ≤ 500.8x 3.2y < 500.8x 3.2y ≤ 50
Mathematics
2 answers:
Amanda [17]4 years ago
8 0

It's the second one

Step-by-step explanation: just did it on edge.

azamat4 years ago
7 0
Given that each book weighs 3.2 pounds, the total weight of x new books is 3.2x. Similarly, the total weight of the new workbooks is 0.8y. Thus, the total weight of the books is 3.2x + 0.8y. This value should be lesser than or equal to 50. The inequality is expressed as, 3.2x + 0.8y ≤ 50. This is second among the choices. 
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Factor 36x + 18 + 12y using GCF HURRY!!!!
worty [1.4K]

Answer:

6x + 3 + 2y

Step-by-step explanation:

The GCF is 6 as thats the largest factor, so you divide the numbers by 6.

6x + 3 + 2y

6 0
3 years ago
Help please :( thank u
kolbaska11 [484]

Answer:

ok

Step-by-step explanation:

8 0
4 years ago
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The plane x + y + 2z = 12 intersects the paraboloid z = x2 + y2 in an ellipse. Find the points on the ellipse that are nearest t
Soloha48 [4]

The distance between a point (x,y,z) and the origin is \sqrt{x^2+y^2+z^2}. But since (\sqrt{f(x)})'=\frac{f'(x)}{2\sqrt{f(x)}}, both f(x) and \sqrt{f(x)} have the same critical points, so we can consider instead the squared distance, x^2+y^2+z^2.

We're looking for the extrema of x^2+y^2+z^2 subject to x+y+2z=12 and z=x^2+y^2. The Lagrangian is

L(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda(x+y+2z-12)+\mu(z-x^2-y^2)

with critical points where the partial derivatives vanish:

L_x=2x+\lambda-2\mu x=0\implies\lambda=2x(\mu-1)

L_y=2y+\lambda-2\mu y=0\implies\lambda=2y(\mu-1)

L_z=2z+2\lambda+\mu=0

L_\lambda=x+y+2z-12=0

L_\mu=z-x^2-y^2=0

From the first two equations, it follows that x=y.

Then in the last two equations,

x+y+2z-12=0\implies x+z=6

z-x^2-y^2=0\implies z=2x^2

\implies x+2x^2=6\implies2x^2+x-6=(2x-3)(x+2)=0\implies x=\dfrac32\text{ or }x=-2

If x=\frac32, then z=6-\frac32=\frac92; if x=-2, then z=8.

So there are two critical points, \left(\frac32,\frac32,\frac92\right) and (-2,-2,8).

Let f(x,y,z)=\sqrt{x^2+y^2+z^2}. We have a minimum distance of f\left(\frac32,\frac32,\frac92\right)=\boxed{\frac{3\sqrt{11}}2} and maximum distance of f(-2,-2,8)=\boxed{6\sqrt2}.

8 0
4 years ago
Use elimination to solve!!!!!!
REY [17]
For easier notation lets use: Nn=x and Nd=y

x+y=22
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6 0
3 years ago
Solve the equation. Round the answer to the nearest tenth.
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X ≈ 4.6

Substitute the value and solve


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