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Zigmanuir [339]
3 years ago
10

Solve the inequality. y−2x ≤ −3

Mathematics
1 answer:
inna [77]3 years ago
4 0
For x, the answer is x >= -(3 - y)/2, for y it's y <= 2x -3.
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Katherine's paycheck was $436.14. She put 1/3 of her paycheck into a savings acount and used 1/2 for what was left to pay for he
d1i1m1o1n [39]

Answer:

$145.38

Step-by-step explanation:

436.14-(1/3)(436.14)-(1/2)(436.14) = 145.38

7 0
3 years ago
Solve for x.<br>-4x - 76 &lt; -52<br>​
Roman55 [17]

First you add 76 to get

-4x<24

And Then you divide by -4 and sincr its a negative you flip the sign to get

x>-6

That might be wrong Idk

4 0
3 years ago
100 points anwser asap
natita [175]

Answer:

23

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
What is the expansion of (3+x)^4
Vlad1618 [11]

Answer:

\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81

Step-by-step explanation:

Considering the expression

\left(3+x\right)^4

Lets determine the expansion of the expression

\left(3+x\right)^4

\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i

a=3,\:\:b=x

=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i

Expanding summation

\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}

i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0

i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1

i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2

i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3

i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4

=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4

=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4

as

\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81

\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x

\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2

\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3

\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4

so equation becomes

=81+108x+54x^2+12x^3+x^4

=x^4+12x^3+54x^2+108x+81

Therefore,

  • \left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81
6 0
3 years ago
I need help on this 6th grade math problem
vivado [14]

Answer: x<17.5

Step-by-step explanation:

Subtract  from both sides: x+2.5-2.5<20-2.5

Simplify the arithmetic: x<20-2.5

Simplify the arithmetic: x<17.5

Hope it helps!

4 0
2 years ago
Read 2 more answers
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