Y=3 - 2x 3x + 4y = 1

Which table can be created using the equation below? –2 + 4x = y A 2-column table with 3 rows. Column 1 is labeled x with entrie

Pachacha [2.7K]

The correct answer is C. A 2-column table with 3 rows. Column 1 is labeled x with entries negative 5, 0, 3. Column 2 is labeled y with entries negative 18, negative 2, 10.

Explanation:

The purpose of an equation is to show the equivalence between two mathematical expressions. This implies in the equation "–2 + 4x = y" the value of y should always be the same that -2 + 4x. Additionally, if a table is created with different values of x and y the equivalence should always be true. This occurs only in the third option.

x y

5 -18

0 -2

3 10

First row:

-2 + 4 (5) = y (5 is the value of x which is first multyply by 4)

-2 + 20 = -18 (value of y in the table)

Second row:

-2 + 4 (0) y

-2 + 0 = -2

Third row:

-2 + 4 (3) = y

-2 + 12 = 10

5 0

3 years ago

Read 2 more answers

Which is equivalent to sqrt x^10 A. X-5 B. X -1/5 C. X sqrt 10 D. X 1/5 E. X^5

Lesechka [4]

<h3>Answer: E) x^5</h3>

$\sqrt{x^{10}} = x^5$

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

Explanation:

We simply take half of the exponent 10 to get 5. This applies to square roots only.

So the rule is $\sqrt{a^b} = a^{b/2}$

A more general rule is

$\sqrt[n]{a^b} = a^{b/n}$

If n = 2, then we're dealing with square roots like with this problem. In this case, a = x and b = 10.

5 0

3 years ago

Read 2 more answers

Vanessa reads door books in seven weeks. How many books did she read each week?

stiks02 [169]

49 books u gotta multiply 7 days times 7 weeks

4 0

3 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$