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2 years ago

13

X-9>6 how do I find the answer

2 answers:

Anvisha [2.4K]2 years ago

5 0

If you would like to solve the inequation X - 9 > 6, you can calculate this using the following steps:

X - 9 > 6 /+9
X - 9 + 9 > 6 + 9
X + 0 > 15
X > 15

The correct result would be X > 15.

Rainbow [258]2 years ago

3 0

Solve for X by adding 9 to each side to undo the subtraction. 6+9=15 so X > 15.

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8 ten thousands 4 thousands 3 hundreds 6 tens 7 ones in Standard form

abruzzese [7]

80,000+4,000+300+60+7

84,367

The standard form is 84,367.

Hope this helps!

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2 years ago

ILL GIVE YOU BRAINLIST !!!! What is the slope of this table?

VLD [36.1K]

Answer: Slope is m = 2/3

Step-by-step explanation:

-You have exactly four points on a table:

$(3,2) (6,4) (9,6) (12,8)$

-Use the slope formula and use any two points from those four points in order to find the slope:

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

-The two points I use are $(3, 2)$ and $(6,4 )$ for the equation:

$m = \frac{4 - 2}{6-3}$

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So, the slope of the table is $\frac{2}{3}$ .

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2 years ago

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BigorU [14]

Answer:

2 fruit for every 3 monsters

Step-by-step explanation:

3 0

2 years ago

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Name the following<br> transformations

Elden [556K]

Answer:

reflection, rotation, and translation.

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2 years ago

Please please please help and solve this with steps, much help needed thank you :) 20 points for this!

leonid [27]

Answer:

$y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}$

(Please vote me Brainliest if this helped!)

Step-by-step explanation:

$\frac{dy}{dx}y=x^3+2x-5x+3$

$\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}$

$\mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)$

1. $\mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:$

$y'\:y=x^3+2x-5x+3$

2. $\mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}$

$yy'\:=x^3-3x+3$

$N\left(y\right)\cdot y'\:=M\left(x\right)$

$N\left(y\right)=y,\:\quad M\left(x\right)=x^3-3x+3$

3. $\mathrm{Solve\:}\:yy'\:=x^3-3x+3:\quad \frac{y^2}{2}=\frac{x^4}{4}-\frac{3x^2}{2}+3x+c_1$

4. $\mathrm{Isolate}\:y:\quad y=\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}}$

5. $\mathrm{Simplify}$

$y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}$

7 0

2 years ago

Read 2 more answers