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3 years ago

Complete the ratio table to convert the units of measure from centimeters to inches or inches to

Mathematics

2 answers:

Pie3 years ago

8 0

Answer:

5.08 centimeters equals 2 inches

10 inches equals 25.4 centimeters

Step-by-step explanation:

we have to complete the following table.

According to the information contained in table 2.54 centimeters it is equal to 1 inch

The first information we have to complete is how many inches equals 5.08 centimeters

To solve it we apply a simple rule of three

$2.54 centimeters \longrightarrow 1 inch \\5.08 centimeters \longrightarrow x$

$x=\frac{(5.08)(1)}{2.54} \\x=2$

5.08 centimeters equals 2 inches

The second information that we have to complete is how many centimeters equals 10 inches

$1 inch \longrightarrow 2.54 centimeters\\10 inches\longrightarrow x \\x=\frac{(10)(2.54)}{1} \\x= 25.4$

10 inches equals 25.4 centimeters

LenaWriter [7]3 years ago

4 0

Answer:

2.45 m in inches = 96.457 in. 2.45 meters to inches = 96.457″. To convert 2.45 meters to inches you have to divide the length expressed in the base unit of length in the International System (SI) of Units by 0.0254 .25.4Centimeters

Step-by-step explanation:

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9966 [12]

Answer:

(b) -m, m + 3

Step-by-step explanation:

x² − 3x − m(m + 3) = 0

x² − 3x = m(m + 3)

x² − 3x + 9/4 = m(m + 3) + 9/4

(x − 3/2)² = m(m + 3) + 9/4

(x − 3/2)² = m² + 3m + 9/4

(x − 3/2)² = (m + 3/2)²

x − 3/2 = ±(m + 3/2)

x − 3/2 = m + 3/2, -m − 3/2

x = m + 3, -m

6 0

3 years ago

Solve the following system of equations using Substitution: Submit your answer as an ordered pair. 2x−3y=32x-3y=32x−3y=3 x=y+2x=

exis [7]

9514 1404 393

Answer:

(3, 1)

Step-by-step explanation:

We assume you want the solution to the system ...

2x−3y=3
x=y+2

The second equation gives a nice expression for x, so we can use that in the first equation.

2(y+2) -3y = 3 . . . . substitute for x in the first equation

2y +4 -3y = 3 . . . . . eliminate parentheses

-y = -1 . . . . . . . . . . . collect terms, subtract 4

y = 1 . . . . . . . . . . . . multiply by -1

x = 1 +2 = 3 . . . . . . substitute for y in the second equation

The solution is (x, y) = (3, 1).

4 0

3 years ago

Calculus 3 help please.

Reptile [31]

I assume each path $C$ is oriented positively/counterclockwise.

(a) Parameterize $C$ by

$\begin{cases} x(t) = 4\cos(t) \\ y(t) = 4\sin(t)\end{cases} \implies \begin{cases} x'(t) = -4\sin(t) \\ y'(t) = 4\cos(t) \end{cases}$

with $-\frac\pi2\le t\le\frac\pi2$ . Then the line element is

$ds = \sqrt{x'(t)^2 + y'(t)^2} \, dt = \sqrt{16(\sin^2(t)+\cos^2(t))} \, dt = 4\,dt$

and the integral reduces to

$\displaystyle \int_C xy^4 \, ds = \int_{-\pi/2}^{\pi/2} (4\cos(t)) (4\sin(t))^4 (4\,dt) = 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt$

The integrand is symmetric about $t=0$ , so

$\displaystyle 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \,dt$

Substitute $u=\sin(t)$ and $du=\cos(t)\,dt$ . Then we get

$\displaystyle 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^1 u^4 \, du = \frac{2^{13}}5 (1^5 - 0^5) = \boxed{\frac{8192}5}$

(b) Parameterize $C$ by

$\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{3^2+4^2} \, dt = 5\,dt$

and

$\displaystyle \int_C x e^y \, ds = \int_0^1 (3t-2) e^{4t} (5\,dt) = 5 \int_0^1 (3t - 2) e^{4t} \, dt$

Integrate by parts with

$u = 3t-2 \implies du = 3\,dt \\\\ dv = e^{4t} \, dt \implies v = \frac14 e^{4t}$

$\displaystyle \int u\,dv = uv - \int v\,du$

$\implies \displaystyle 5 \int_0^1 (3t-2) e^{4t} \,dt = \frac54 (3t-2) e^{4t} \bigg|_{t=0}^{t=1} - \frac{15}4 \int_0^1 e^{4t} \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} e^{4t} \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} (e^4 - 1) = \boxed{\frac{5e^4 + 55}{16}}$

$\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{(-2)^2 + 1^2 + 3^2} \, dt = \sqrt{14} \, dt$

and

$\displaystyle \int_C y^2 z \, ds = \int_0^1 (t+1)^2 (3t+2) \left(\sqrt{14}\,ds\right) \\\\ ~~~~~~~~ = \sqrt{14} \int_0^1 \left(3t^3 + 8t^2 + 7t + 2\right) \, dt \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 t^4 + \frac83 t^3 + \frac72 t^2 + 2t\right) \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 + \frac83 + \frac72 + 2\right) = \boxed{\frac{107\sqrt{14}}{12}}$