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3 years ago

Name the Property of Equality that justifies this statement:

1 answer:

5 0

Symmetric because it is the same thing and also i really dont know so dont hmu im just tryna get points lol sorry for being garabage

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Which ordered pair is generated from the equation shown below? y = 2x + 4 A. (1, 6) B. (1, 2) C. (3, 6) D. (8, 16)

USPshnik [31]

Answer:

The ordered pair generated from the equation is (1, 6).

Step-by-step explanation:

An ordered pair is a pair of numbers, representing two variables, in a specific order. For instance, (x, y) = (1, 2) here x = 1 and y = 2.

The equation provided is:

$y=2x+4$

Check for all the options:

A (1, 6): $y=2x+4\Rightarrow\ 6=(2\times 1) +4\Rightarrow\ 6=6$
B (1, 2): $y=2x+4\Rightarrow\ 2=(2\times 1) +4\Rightarrow\ 2\neq 6$
C (3, 6): $y=2x+4\Rightarrow\ 6=(2\times 3) +4\Rightarrow\ 6\neq10$
D (8, 16): $y=2x+4\Rightarrow\ 16=(2\times 8) +4\Rightarrow\ 16\neq20$

Thus, the ordered pair generated from the equation is (1, 6).

8 0

3 years ago

A line passes through the points (1, 4) and (3,-4). Which is the equation of the line?

andrew-mc [135]

Answer:

Y=-4x+8

Step-by-step explanation:

3 0

2 years ago

-(9-15)+(-8+2)-(24-30)=

AleksAgata [21]

The answer is -6
(9-15)=-6
(-8+2)=-6
(24-30)=-6
-(-6)+(-6)-(-6)= -6

3 0

2 years ago

Given g(x) = -x + 3, solve for x when g(x) = -1.

cluponka [151]

Answer:

Step-by-step explanation:

It's ok! This is simple. Simply plug in your given values. Since g(x) = -1, plug that into the equation.

g(x) = -x + 3

g(-1) = -(-1) + 3

=1+3

7 0

2 years ago

Evaluate the surface integral. s y ds, s is the helicoid with vector equation r(u, v) = u cos(v), u sin(v), v , 0 ≤ u ≤ 6, 0 ≤ v

Juliette [100K]

Compute the surface element:

$\mathrm dS=\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv$

$\vec r(u,v)=(u\cos v,u\sin v,v)\implies\begin{cases}\vec r_u=(\cos v,\sin v,0)\\\vec r_v=(-u\sin v,u\cos v,1)\end{cases}$

$\|\vec r_u\times\vec r_v\|=\sqrt{\sin^2v+(-\cos v)^2+u^2}=\sqrt{1+u^2}$

So the integral is

$\displaystyle\iint_Sy\,\mathrm dS=\int_0^\pi\int_0^6u\sin v\sqrt{1+u^2}\,\mathrm du\,\mathrm dv$

$=\displaystyle\left(\int_0^\pi\sin v\,\mathrm dv\right)\left(\int_0^6u\sqrt{1+u^2}\,\mathrm du\right)$

$=\dfrac23(37^{3/2}-1)$

4 0

2 years ago