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3 years ago

X − 3 ≥ 6 pls help me

Mathematics

2 answers:

Fantom [35]3 years ago

6 0

Answer:

x $\geq$ 9

Step-by-step explanation: i took a quiz with this on it

juin [17]3 years ago

5 0

Answer:

x ≥ 9

Step-by-step explanation:

Solve it like any equation

Add 3 to both sides

x ≥ 9

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house numbers on one side of a street are consecutive even integers. the sum of the first three house numbers on the street is 6

Vanyuwa [196]

Consecutive even integers..x, x + 2, x + 4

x + (x + 2) + (x + 4) = 6216
3x + 6 = 6216
3x = 6216 - 6
3x = 6210
x = 6210/3
x = 2070

x + 2 = 2070 + 2 = 2072
x + 4 = 2070 + 4 = 2074 <===house 3

4 0

3 years ago

Haven't run across one with three terms. help? preferably explain how you got the answer too!!

const2013 [10]

Answer:

c^3 + c^2 - 7c + 20

Step-by-step explanation:

First, expand the expression using distributive property.

c^2(c+4) - 3c(c+4) + 5(c+4)

c^3 + 4c^2 - 3c^2 - 12c + 5c + 20

Lastly, simplify like terms.

c^3 + c^2 - 7c + 20

5 0

3 years ago

Describe the effect of reflecting EFGH over the x-axis to E'F'G'H'.

telo118 [61]

Answer:D. The sign of each y- coordinate change.

Step-by-step explanation: because everything on the X axis stayed the same but on the Y axis it decreased

5 0

3 years ago

Read 2 more answers

I need help quickly plz!!!!

natima [27]

Answer:

25:3

Step-by-step explanation:

50 / 6 is simplified to 25 / 3 which would equal 25 : 3.

7 0

3 years ago

Let C be the curve of intersection of the parabolic cylinder x^2 = 2y, and the surface 3z = xy. Find the exact length of C from

Maslowich

I've attached a plot of the intersection (highlighted in red) between the parabolic cylinder (orange) and the hyperbolic paraboloid (blue).

The arc length can be computed with a line integral, but first we'll need a parameterization for $C$ . This is easy enough to do. First fix any one variable. For convenience, choose $x$ .

Now, $x^2=2y\implies y=\dfrac{x^2}2$ , and $3z=xy\implies z=\dfrac{x^3}6$ . The intersection is thus parameterized by the vector-valued function

$\mathbf r(x)=\left\langle x,\dfrac{x^2}2,\dfrac{x^3}6\right\rangle$

where $0\le x\le 4$ . The arc length is computed with the integral

$\displaystyle\int_C\mathrm dS=\int_0^4\|\mathbf r'(x)\|\,\mathrm dx=\int_0^4\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}\,\mathrm dx$

Some rewriting:

$\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}=\sqrt{\dfrac{x^2}{36}}\sqrt{x^4+9x^2+36}=\dfrac x6\sqrt{x^4+9x^2+36}$

Complete the square to get

$x^4+9x^2+36=\left(x^2+\dfrac92\right)^2+\dfrac{63}4$

So in the integral, you can substitute $y=x^2+\dfrac92$ to get

$\displaystyle\frac16\int_0^4x\sqrt{\left(x^2+\frac92\right)^2+\frac{63}4}\,\mathrm dx=\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy$

Next substitute $y=\dfrac{\sqrt{63}}2\tan z$ , so that the integral becomes

$\displaystyle\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy=\frac{21}{16}\int_{\arctan(3/\sqrt7)}^{\arctan(41/(3\sqrt7))}\sec^3z\,\mathrm dz$

This is a fairly standard integral (it even has its own Wiki page, if you're not familiar with the derivation):

$\displaystyle\int\sec^3z\,\mathrm dz=\frac12\sec z\tan z+\frac12\ln|\sec x+\tan x|+C$

So the arc length is

$\displaystyle\frac{21}{32}\left(\sec z\tan z+\ln|\sec x+\tan x|\right)\bigg|_{z=\arctan(3/\sqrt7)}^{z=\arctan(41/(3\sqrt7))}=\frac{21}{32}\ln\left(\frac{41+4\sqrt{109}}{21}\right)+\frac{41\sqrt{109}}{24}-\frac98$

4 0

4 years ago