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3 years ago

Can someone help me please

Mathematics

2 answers:

mestny [16]3 years ago

6 0

You got it correct !

Rudik [331]3 years ago

5 0

Answer:

its correct don't sweat

Step-by-step explanation:

If I am right you are trying to find where the line cross each other it runs on 4 and rises to 2.

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If G = {(-1, 7),(-8, 2),(0, 0),(6, 6)), then the range of G is

lana66690 [7]

Answer:

$\{0, 2, 6, 7\}$

Step-by-step explanation:

You are given the relation $G = \{(-1, 7),(-8, 2),(0, 0),(6, 6)\}$

This means that

$f(-1)=7\\ f(-8)=0\\f(0)=0\\f(6)=6$

The input values of this relation are $-1,\ -8,\ 0,\ 6$ , so the domain of this relation is $\{-8,-1,0,6\}$

The output values of this relation are $7,\ 2,\ 0,\ 6$ , so the range of this relation is $\{0, 2, 6, 7\}$

8 0

3 years ago

Ashton has an offer to buy an item with a sticker price of $4900 by paying

Alla [95]

B
The correct answer is b

6 0

3 years ago

Homework ree

umka21 [38]

Answer:

325 miles and 77 dollars

Step-by-step explanation:

First, create equations. Let x equal the miles driven. The first plan expression is 38 + 0.12x and the second plan expression is 51 + 0.08x. Setting these equal and solve for x: 38 + 0.12x = 51 + 0.08x -> 0.04x = 13 -> x = 325 miles.

The cost is found by substituting it into any of the expressions: 51 + 0.08(325) = 77 or 38 + 0.12(325) = 77

6 0

3 years ago

Please help me. Hurry, linear equations.

Lisa [10]

Answer:

Let's simplify step-by-step.

3−3/4x−1/4x−7

=3+ −3/4x+ −1/4x+ −7

Combine Like Terms:

=3+ −3/4x+ −1/4x+ −7

=(−3/4x+ −1/4x)+(3+ −7)

=−x+ −4

Answer:=−x−4

Step-by-step explanation:

answer:-4

8 0

3 years ago

Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)

Triss [41]

$g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}$

The surface element is

$\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

and the integral is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)$

###

To compute the last integral, you can integrate by parts:

$u=r\implies\mathrm du=\mathrm dr$

$\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}$

$\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr$

For this integral, consider a substitution of

$r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds$

$\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds$

$\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds$

$=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds$

and the result above follows.

4 0

4 years ago