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3 years ago

PLEASE HURRY!!!

Mathematics

1 answer:

Gelneren [198K]3 years ago

4 0

Answer:

The third option, x = 2 over 3.

Step-by-step explanation:

x + 4/3 = 2

x = 2 - 1 1/3

x = 2/3

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2 years ago

A line passes through the points (5, –7.5) and has a y-intercept of 10.

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3 years ago

Which ordered pair represents the solution to the system of equations? {−5x+7y=−196x−2y=10

k0ka [10]

Answer:
x = 1
y = -2

Explanation:
First, solve for x by subtracting 7y from both sides of the first equation and then divide both sides by -5.

-5x + 7y = -19
-5x + 7y - 7y = -19 - 7y
-5x = -19 - 7y
-5x/ -5 = -19 - 7y/ -5
x = -19 - 7y/ -5

Next, substitute the value of x into the second equation and solve.

6 - (-19 - 7y)/ 5 - 2y = 10
- 6(-19 - 7y)/ 5 - 2y = 10

Then, solve for y

- 6(-19 - 7y)/ 5 - 2y = 10
-6(-19 - 7y) - 10y = 50
114 + 42y - 10y = 50
144 + 32y = 50
32y = 50 - 144
32y = -64
32y/ 32 = -64/ 32
y = -2

Finally, substitute the value of y into the x value we found earlier and solve

x = - 19 - 7y/ 5
x = - 19 - 7 • -2/ 5
x = 1

8 0

3 years ago

The line integral of (2x+9z) ds where the curve is given by the parametric equations x=t, y=t^2, z=t^3 for t between 0 and 1. Pl

Naya [18.7K]

Let r = (t,t^2,t^3)

Then r' = (1, 2t, 3t^2)

General Line integral is:
$\int_a^b f(r) |r'| dt$

The limits are 0 to 1
f(r) = 2x + 9z = 2t +9t^3
|r'| is magnitude of derivative vector $\sqrt{(x')^2 + (y')^2 + (z')^2}$

$\int_0^1 (2t+9t^3) \sqrt{1+4t^2 +9t^4} dt$

Fortunately, this simplifies nicely with a 'u' substitution.

Let u = 1+4t^2 +9t^4

du = 8t + 36t^3 dt

$\int_0^1 \frac{2t+9t^3}{8t+36t^3} \sqrt{u} du \\ \\ \int_0^1 \frac{2t+9t^3}{4(2t+9t^3)} \sqrt{u} du \\ \\ \frac{1}{4} \int_0^1 \sqrt{u} du$

After integrating using power rule, replace 'u' with function for 't' and evaluate limits:
$=\frac{1}{4} |_0^1 (\frac{2}{3}) (1+4t^2 +9t^4)^{3/2} \\ \\ =\frac{1}{6} (14^{3/2} - 1)$

7 0

3 years ago