<img src="https://tex.z-dn.net/?f=6%20%3C%20%20-%205t%20-%204" id="TexFormula1" title="6 < - 5t - 4" alt="6 < - 5t

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3241004551 [841]

3 years ago

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Mathematics

1 answer:

LekaFEV [45]3 years ago

6 0

Answer:

Step-by-step explanation: We would solve this exactly like we would a variable equation. If it makes it easier imagine there is equal sign instead of a inequality sign.

First we add 4 to both sides.

6<-5t-4

10<-5t

Then we divide both sides by -5.

10<-5t

10/-5<-5t/-5

The two -5s cancel out only leaving the 10 to be divided by the -5......

but a law in math tells us if we divide in math by a negative or multiply by one then we must reverse the inequality sign.

.....which leaves us with the answer.

-2>t

Hope this helps!

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1) $(3x+2)+(4-5x)=-2x+6$

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Step-by-step explanation:

In mathematics there are rules related to complex numbers, specifically in the case of addition and multiplication:

Addition:

If we have two complex numbers written in their binomial form, the sum of both will be a complex number whose real part is the sum of the real parts and whose imaginary part is the sum of the imaginary parts (similarly as the sum of two binomials).

For example, the addition of these two binomials is:

$(3x+2)+(4-5x)=3x-5x+2+4=-2x+6$

Similarly, the addition of two complex numbers is:

$(3+2i)+(4-5i)=3+4+2i-5i=7-3i$ Here the complex part is the number with the $i$

Multiplication:

If we have two complex numbers written in their binomial form, the multiplication of both will be the same as the multiplication (product) of two binomials, taking into account that $i^{2}=-1$ .

For example, the multiplication of these two binomials is:

$(1-3x)(2+x)=2+x-6x-3x^{2}=-3x^{2}-5x+2$

Similarly, the multiplication of two complex numbers is:

$(1-3i)+(2+i)=2+i-6i-3i^{2}=2-5i-3(-1)=5-5i$

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