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wariber [46]
3 years ago
12

Numbers greater than -6

Mathematics
2 answers:
Masteriza [31]3 years ago
8 0
-4 -3 -2 -1 , 0 , 1
MakcuM [25]3 years ago
3 0
 -5,-4,-3,-2,-1,0,1,2 etc
You might be interested in
Simplify (12a5−6a−10a3)−(10a−2a5−14a4) . Write the answer in standard form
almond37 [142]

Answer:

=14a^5+14a^4-10a^3-16a

Step-by-step explanation:

\left(12a^5-6a-10a^3\right)-\left(10a-2a^5-14a^4\right)\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=12a^5-6a-10a^3-\left(10a-2a^5-14a^4\right)\\-\left(10a-2a^5-14a^4\right):\quad -10a+2a^5+14a^4\\-\left(10a-2a^5-14a^4\right)\\\mathrm{Distribute\:parentheses}\\=-\left(10a\right)-\left(-2a^5\right)-\left(-14a^4\right)\\Apply\:minus-plus\:rules\\-\left(-a\right)=a,\:\:\:-\left(a\right)=-a\\=-10a+2a^5+14a^4\\=12a^5-6a-10a^3-10a+2a^5+14a^4

\mathrm{Simplify}\:12a^5-6a-10a^3-10a+2a^5+14a^4:\quad 14a^5+14a^4-10a^3-16a12a^5-6a-10a^3-10a+2a^5+14a^4\\Group\:like\:terms\\=12a^5+2a^5+14a^4-10a^3-6a-10a\\\mathrm{Add\:similar\:elements:}\:12a^5+2a^5=14a^5\\=14a^5+14a^4-10a^3-6a-10a\\\mathrm{Add\:similar\:elements:}\:-6a-10a=-16a\\=14a^5+14a^4-10a^3-16a

4 0
2 years ago
Solve 4 brainliest xx
serious [3.7K]
Y=6x+4
Blanks from left to right:
-14,-2,10,22,6x+4
5 0
2 years ago
Need help quick please
valina [46]

Answer:

Disagree

Step-by-step explanation:

Remember rotating 90 degrees  means (y,-x)

A= (1,2)

A'= (2,-1)

So it is disagreed.

Hope this help :)

3 0
2 years ago
Read 2 more answers
Given: y" - 2y' = 6t + 5e^2t. Find the correct form to use for y_p if the equation is solved using Undetermined coefficients. Do
const2013 [10]

Answer:

y_p=A+Bt+Ce^{2t}

Step-by-step explanation:

Given: y'' - 2y' = 6t + 5e^{2t}.

we need to find the correct form for y_p if the equation is solve using undetermined coefficients.

A first order differential equation \frac{\mathrm{d} y}{\mathrm{d} x}=f\left ( x,y \right ) is said to be homogeneous if f(tx,ty)=f(x,y) for all t.

Consider homogeneous equation y''-2y'=0

Let y=e^{rt} be the solution .

We get (r^2-2r)e^{rt}=0

Since e^{rt}\neq 0, r^2-2r=0.

So, we get solution as y_c=c_1+c_2e^{2t}

As constant term and e^{2t} are already in the R.H.S of equation

y" - 2y' = 6t + 5e^{2t}, we can take y_p as y_p=A+Bt+Ce^{2t}

6 0
3 years ago
Find the domains of the following functions:
larisa86 [58]
I think this is the answer to that,,, >y = 5x^2
3 0
2 years ago
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