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3 years ago

12

2x+y=5 -x+2y=4 How do you solve this

1 answer:

d1i1m1o1n [39]3 years ago

7 0

The answer to your problem is
x=5/2
y=5

x=-4
y=2

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A sporting goods store sold 2.5 times as many footballs as basketballs last year. Which statement is true?

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3 years ago

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Jim sells used drones for 70% of the original manufacturer's purchase price of a new drone. Jim also sells new drones for 15% mo

adell [148]

The function for the amount saved if they purchase a used drone rather than a new drone from Jim's store is f(x) = 0.45x

He sells used drones at 70% of the price (x).

So, the selling price of a used drone is:

$\mathbf{Used = 70\% \times x}$

Express percentage as decimal

$\mathbf{Used = 0.7 \times x}$

$\mathbf{Used = 0.7x}$

Also, he sells new drones at 15% more of the price (x).

So, the selling price of a new drone is:

$\mathbf{New= (1 + 15\%) \times x}$

Express percentage as decimal

$\mathbf{New= (1 + 0.15) \times x}$

$\mathbf{New= 1.15 \times x}$

$\mathbf{New= 1.15x}$

The amount saved is the difference between the prices of the drones.

So, we have:

$\mathbf{Saved =New - Old}$

This gives

$\mathbf{Saved =1.15x- 0.7x}$

Subtract

$\mathbf{Saved =0.45x}$

Express as a function

$\mathbf{f(x) = 0.45x}$

Hence, the function for the amount saved is f(x) = 0.45x

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2 years ago

a) find center of mass of a solid of constant density bounded below by the paraboloid z=x^2+y^2 and above by the plane z=4.(b) F

slava [35]

Answer: x-bar = y-bar = 0 whereas z-bar = 8/3

M= (c^2)/8 which is intern equal to 2 $\sqrt{2}$

Step-by-step explanation:

Find the area, by setting the limits as

= $4\cdot \int _0^{\frac{\pi }{2}}\int _0^2\int _{r^2}^4\:rdzdrd\theta$

$=4\cdot \int _0^{\frac{\pi }{2}}\int _0^2r\cdot \:4-r^3drd\theta$

$=4\cdot \int _0^{\frac{\pi }{2}}4d\theta$

$=8\pi$

Therefore;

Mxy= $\int _0^{2\pi }\int _0^2\int _{r^2}^4\:zrdzdrd\theta$

z-bar = 8/3

M= 8 $\pi$ dividing it into two volume gives us = 4 $\pi$

means $4\pi =\int _0^{2\pi }\int _0^{\sqrt{c}}\int _r^c\:rdzdrd\theta$

$4\pi =\left(\pi c^2-2\pi \frac{c^{\frac{3}{2}}}{3}\right)$

c=2 $\sqrt{2}$

5 0

3 years ago