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3 years ago

Solve for and in the following simultaneous equations below; 5x−4y=8 and 2x+y=11.

Mathematics

2 answers:

lorasvet [3.4K]3 years ago

7 0

Answer:

x=4 and y=3

Step-by-step explanation:

put one of the equations in slope form and solve

DanielleElmas [232]3 years ago

3 0

Answer:x=4 y=3

Step-by-step explanation:

5x-4y=8.............(1)

2x+y=11................(11)

From (11) y=11-2x

Substitute y=11-2x in 5x-4y=8

5x-4y=8

5x-4(11-2x)=8

Open bracket

5x-44+8x=8

Collect like terms

5x+8x=8+44

13x=52

Divide both sides by 13

13x/13=52/13

x=4

Substitute x=4 in 2x+y=11

2(4)+y=11

8+y=11

Collect like terms

y=11-8

y=3

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Help!! Find missing value for each quadratic function

Svetllana [295]

I'll do the first one to get you started. The answer is -3

=================================================

Explanation:

When x = -1, y = 0 as shown in the first column. The second column says that (x,y) = (0,1) and the third column says (x,y) = (1,0)

We'll use these three points to determine the quadratic function that goes through them all.

The general template we'll use is y = ax^2 + bx + c

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Plug in (x,y) = (0,1). Simplify

y = ax^2 + bx + c

1 = a*0^2 + b*0 + c

1 = 0a + 0b + c

1 = c

c = 1

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Plug in (x,y) = (-1,0) and c = 1

y = ax^2 + bx + c

0 = a*(-1)^2 + b(-1) + 1

0 = 1a - 1b + 1

a-b+1 = 0

a-b = -1

a = b-1

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Plug in (x,y) = (1,0), c = 1, and a = b-1

y = ax^2 + bx + c

0 = a(1)^2 + b(1) + 1 ... replace x with 1, y with 0, c with 1

0 = a + b + 1

0 = b-1 + b + 1 ... replace 'a' with b-1

0 = 2b

2b = 0

b = 0/2

b = 0

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If b = 0, then 'a' is...

a = b-1

a = 0-1

a = -1

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In summary so far, we found: a = -1, b = 0, c = 1

Therefore, y = ax^2 + bx + c turns into y = -1x^2 + 0x + 1 which simplifies to y = -x^2 + 1

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The last thing to do is plug x = 2 into this equation and simplify

y = -x^2 + 1

y = -2^2 + 1

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3 years ago

The median rainfall during a spring storm in the lowlands is about 4 mm less than a spring storm in the highlands. The mean rain

butalik [34]

Median is just the value that is neither the highest nor the lowest compared with the other values in the set

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3 years ago

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Use polar coordinates to find the volume of the given solid. Inside both the cylinder x2 y2 = 1 and the ellipsoid 4x2 4y2 z2 = 6

Anton [14]

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

V= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

<h3>What is Volume of Solid in polar coordinates?</h3>

To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.

Consider the cylinder, $x^{2}+y^{2} =1$ and the ellipsoid, $4x^{2}+ 4y^{2} + z^{2} =64$

In polar coordinates, we know that

$x^{2}+y^{2} =r^{2}$

So, the ellipsoid gives

$4{(x^{2}+ y^{2)} + z^{2} =64$

4( $r^{2}$ ) + $z^{2}$ = 64

$z^{2}$ = 64- 4( $r^{2}$ )

z=± $\sqrt{64-4r^{2} }$

So, the volume of the solid is given by:

V= $\int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta$

= $2\int\limits^{2\pi}_ 0 \int\limits^1_0 \, r\sqrt{64-4r^{2} } r dr d\theta$

To solve the integral take, $64-4r^{2}$ = t

dt= -8rdr

rdr = $\frac{-1}{8} dt$

So, the integral $\int\ r\sqrt{64-4r^{2} } rdr$ become

= $\int\ \sqrt{t } \frac{-1}{8} dt$

= $\frac{-1}{12} t^{3/2}$

= $\frac{-1}{12} (64-4r^{2}) ^{3/2}$

so on applying the limit, the volume becomes

V= $2\int\limits^{2\pi}_ {0} \int\limits^1_0{} \, \frac{-1}{12} (64-4r^{2}) ^{3/2} d\theta$

= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta$

V = $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Since, further the integral isn't having any term of $\theta$ .

we will end here.

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Learn more about Volume in polar coordinate here:

brainly.com/question/25172004

#SPJ4

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