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3 years ago

What is the solution of the system? y = 13x - 1 ; y = 9x + 4

Mathematics

1 answer:

Temka [501]3 years ago

6 0

Answer:

$(\frac{5}{4},\frac{61}{4}$ )

Step-by-step explanation:

This system we can solve by substitution method :

y=13x-1

y=9x+4

_______

y=13x-1

13x-1=9x+4 (instead y in second equation we use 13x-1 and first rewrite)

_______

y=13x-1

13x-9x=4+1 (subtract 9x and add 1 for each side)

______

y=13x-1

4x=5 (divide by 4)

_____

y=13x-1

$x=\frac{5}{4}$ (instead x in first eq we write 5/4)

______

$y=13*\frac{5}{4} -1=\frac{65-4}{4}=\frac{61}{4}$

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The scale on a map shows that 200 miles is represen by 5 inches on a map. Which proportion can be used to find the distance on t

Mandarinka [93]

Answer:

1.25 inches

Step-by-step explanation:

200 miles/ 50 miles = 5 inch / x

x = (50 x 5) / 200 = 250 / 200 = 1.25

4 0

4 years ago

Find the mass and the center of mass of a wire loop in the shape of a helix (measured in cm: x = t, y = 4 cos(t), z = 4 sin(t) f

Sholpan [36]

Answer:

Mass

$\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)$

Center of mass

Coordinate x

$\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Coordinate y

$\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Coordinate z

$\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Step-by-step explanation:

Let W be the wire. We can consider W=(x(t),y(t),z(t)) as a path given by the parametric functions

x(t) = t

y(t) = 4 cos(t)

z(t) = 4 sin(t)

for 0 ≤ t ≤ 2π

If D(x,y,z) is the density of W at a given point (x,y,z), the mass m would be the curve integral along the path W

$m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt$

The density D(x,y,z) is given by

$D(x,y,z)=x^2+y^2+z^2=t^2+16cos^2(t)+16sin^2(t)=t^2+16$

on the other hand

$||W'(t)||=\sqrt{1^2+(-4sin(t))^2+(4cos(t))^2}=\sqrt{1+16}=\sqrt{17}$

and we have

$m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt=\\\\\sqrt{17}\displaystyle\int_{0}^{2\pi}(t^2+16)dt=\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)$

The center of mass is the point $(\bar x,\bar y,\bar z)$

where

$\bar x=\displaystyle\frac{1}{m}\displaystyle\int_{W}xD(x,y,z)\\\\\bar y=\displaystyle\frac{1}{m}\displaystyle\int_{W}yD(x,y,z)\\\\\bar z=\displaystyle\frac{1}{m}\displaystyle\int_{W}zD(x,y,z)$

We have

$\displaystyle\int_{W}xD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}t(t^2+16)dt=\\\\=\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)$

$\bar x=\displaystyle\frac{\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

$\displaystyle\int_{W}yD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}4cos(t)(t^2+16)dt=\\\\=16\sqrt{17}\pi$

$\bar y=\displaystyle\frac{16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

$\displaystyle\int_{W}zD(x,y,z)=4\sqrt{17}\displaystyle\int_{0}^{2\pi}sin(t)(t^2+16)dt=\\\\=-16\sqrt{17}\pi$

$\bar z=\displaystyle\frac{-16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

3 0

3 years ago

Pls help me no links I am serious

kykrilka [37]

Answer:

March

Step-by-step explanation:

It’s March because in February there were 45 or so students, but then in March it went all the way up to 120. So, 120 - 45 = 75, which is the most amount of students increasing in a month.

I hope this helped and it wasn’t so confusing!!

8 0

3 years ago

Solve this inequality for x: -46 − 8x > 22.

sveticcg [70]

Answer:

-46-8x>22

22-46=-24

8x/8 -24/8

x=-3

Step-by-step explanation:

7 0