8x-6x-3 what do you do when there are two multiplication signs in multiplicating integers?

2 answers:

5 0

A math sentence involving multiple operations is evaluated according to the Order of Operations. Multiplication is done left to right.

(8) × (-6) × (-3)

= ((8) × (-6)) × (-3)

= (-48) × (-3)

= 144

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The associative and commutative properties of multiplication allow you to rearrange this to any convenient order. You can also remove factors of -1 from the negative numbers and deal with those separately, if you like.

8 × -6 × -3

= 8 × -1 × 6 × -1 × 3

= (-1)×(-1) × 8 × 6 × 3

= 1 × 8 × 6 × 3

= 8 × 6 × 3 . . . . . . 1 times anything is that thing. That's the multiplicative identity property.

= 3 × 6 × 8 . . . . . if you want. (Order for multiplication of numbers doesn't matter. It does matter when multiplying certain other things, like matrices or vectors.)

Wewaii [24]3 years ago

4 0

Those are variable not multiplication signs
Solve the variables and see what the answer is I don't believe you can go any farther than it being 2x-3 unless you want a fraction.

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notsponge [240]

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8 0

3 years ago

Read 2 more answers

What is the seventh term of (x + 4)8?

Arisa [49]

The (p+1)-th term of the Newton binomial expansion

$(a+b)^{n}$

is given by

$t_{p+1}=\dbinom{n}{p}\,a^{n-p}\,b^{p}$
__________________________

We want the 7th term. Hence, we set $p+1$ to be $7:$

$p+1=7~~\Rightarrow~~p=6$

Then, the 7th term is

$t_{7}=\dbinom{8}{6}\,x^{8-6}\,4^{6}\\\\\\ t_{7}=\dfrac{8!}{6!\cdot (8-6)!}\cdot x^{8-6}\cdot 4^{6}\\\\\\ t_{7}=\dfrac{8\cdot 7\cdot \diagup\!\!\!\! 6!}{\diagup\!\!\!\! 6!\cdot 2!}\cdot x^{2}\cdot 4^{6}\\\\\\ t_{7}=\dfrac{8\cdot 7}{2\cdot 1}\cdot x^{2}\cdot 4^{6}\\\\\\ t_{7}=28\cdot x^{2}\cdot 4,096\\\\ \boxed{\begin{array}{c} t_{7}=114,688\,x^{2} \end{array}}$

Correct answer: $\text{A. }114,688\,x^{2}.$

8 0

4 years ago

2+2x=3+2x explain why the equation does not have a solution

cupoosta [38]

Subtracting $2x$ from both sides gives us the false statement $2=3,$ so this equation can never be true and thus has no solutions.