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1 year ago

Write an inequality for the graph. Then, in words, describe all the values of x that make inequality true.

Mathematics

1 answer:

GarryVolchara [31]1 year ago

5 0

The equivalent inequality for the number line is x < 8
The equivalent inequality for the number line is x ≥-5

<h3>Inequality functions </h3>

The inequality expression is an expression not separated by an equal sign,

11) The equivalent inequality for the number line is x < 8. This shows that the values of x that make it true must be less than 8.

Note that the sign is less than because the circle is open and moves towards the left of the number line

11) The equivalent inequality for the number line is x ≥-5. This shows that the values of x that make it true must be greater than or equal to 8.

Note that the sign is greater than or equal to because the circle is closed and moves towards the right of the number line

Learn more on number line here:brainly.com/question/24644930

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

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

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$\left(3+x\right)^4$

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

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Expanding summation

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

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

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

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

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

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

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

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

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

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

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

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

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

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<em>First.</em> Let us prove that the sum of three consecutive integers is divisible by 3.

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