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3 years ago

Factor this expression completely. mr + ns - nr - ms

1 answer:

3 0

Greetings!

$mr+ns-nr-ms$
Rearrange.
$mr-nr-ms+ns$
Factor by grouping.
$r(m-n)+s(-m+n)$
Multiply the expression by -1.
$r(m-n)-1(+s(-m+n))$
Simplify.
$r(m-n)-s(m-n)$
Find the GCF.
$(m-n)(r-s)$

Hope this helps.
-Benjamin

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Find the output , h , when the input , x , is -18 h= 17+x/6

pantera1 [17]

Once you plug in -18 as x you divide it with 6 and the answer of that is -3 then you add 17 and -3 and your output will be 14!

3 0

3 years ago

30 points for whoever helps George Washington...please

faust18 [17]

Answer:

Step-by-step explanation:

A is equal to 5d-5

I just went through all eh choices

It can’t be b or c becuase when distributed, it automatically is -5d+25 or vice versa with a different sign so their out

D is impossible becuase you can’t jsut flip a subtraction problem

And A is possible because adding a negative basically means subtracting so if flipped The signs have to change

8 0

2 years ago

Juan invest $3700 in a simple interest account at a rate of 4% for 15 years

OleMash [197]

Question:

Juan Invest $3700 In A Simple Interest Account At A Rate Of 4% For 15 Years. How Much Money Will Be In The Account After 15 Years?

Answer:

There will be $ 5920 in account after 15 years

Solution:

The simple interest is given by formula:

$S.I = \frac{p \times n \times r }{100}$

Where,

p is the principal

n is number of years

r is rate of interest

From given,

p = 3700

r = 4 %

t = 15 years

Therefore,

$S.I = \frac{3700 \times 4 \times 15 }{100}\\\\S.I = 37 \times 4 \times 15\\\\S.I = 2220$

How Much Money Will Be In The Account After 15 Years?

Total money = principal + simple interest

Total money = 3700 + 2220

Total money = 5920

Thus there will be $ 5920 in account after 15 years

8 0

3 years ago

Which of the following expressions are equivalent? Justify your reasoning,

marshall27 [118]

Answer:

B and D

Step-by-step explanation:

Anything to a negative power means that it is 1/that to the positive power.

E.g. x^-1 = 1/x^1

In other words, anything to the power of a negative switches sides of a fraction (i.e. if in numerator moves to denominator and vice versa.)

1/x^-1 = 1/1/x^1 which is just equal to x, because there are x number of 1/xs in one (1/x * x =1) Therefore Option B is equal to just x.

D: (assuming the first given term is x^1/3 and not X1/3 (?) Correct me if I'm wrong).

x^1/3 * x^1/3 * x^1/3 is also equal to just x.

This is because when multiplying together terms with the same base (x in this case) the exponents just add together, so:

x^1/3 * x^1/3 * x^1/3 = x^(1/3 +1/3 +1/3) = x^1 = x.

Therefore B and D are equivalent because they both equal x.

Hope this helped!

7 0

3 years ago

Can someone please help me with my maths question

DIA [1.3K]

Answer:

$a. \ \dfrac{625 \cdot m}{27 \cdot n^{11}}$

$b. \ \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

Step-by-step explanation:

The question relates with rules of indices

(a) The give expression is presented as follows;

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}$

By expanding the expression, we get;

$\dfrac{m^3 \times n^{-8} \times 5^4 \times m^4}{\left 3^3 \times m^6 \times n^3}$

Collecting like terms gives;

$\dfrac{m^{(3 + 4 - 6)} \times 5^4}{ 3^3 \times n^{3 + 8}} = \dfrac{625 \cdot m}{27 \cdot n^{11}}$

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}= \dfrac{625 \cdot m}{27 \cdot n^{11}}$

(b) The given expression is presented as follows;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \div (x \cdot y^n)^4$

Therefore, we get;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n}$

Collecting like terms gives;

$x^{3 \cdot m + 2 - 4} \times \left (y^{3 \cdot n - 3 -4 \cdot n}} \right ) = x^{3 \cdot m - 2} \times \left (y^{ - 3 -n}} \right ) = x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right )$

$x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right ) = \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n} =\dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

4 0

3 years ago