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2 years ago

The following steps are used to rewrite the polynomial expression (x+2y)(x-5y)

Mathematics

2 answers:

neonofarm [45]2 years ago

7 0

(x+2y)(x-5y)
do foil method:
x^2-5xy+2xy-10y^2
simplify x^2-3xy-10y^2,
so d is correct.

marshall27 [118]2 years ago

4 0

Answers
Step 1) Distributive property
Step 2) Distributive property
Step 3) Commutative property of multiplication
Step 4) Combine like terms

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The distributive property is the rule
a*(b+c) = a*b+a*c

we can make it a bit more complicated if we replace 'a' with some other expression, say x+y.

So
a*(b+c) = a*b + a*c
turns into
(x+y)*(b+c) = (x+y)*b + (x+y)*c
and then we can use the distributive property again to further distribute the outer letters into each parenthesis grouping
(x+y)*(b+c) = (x+y)*b + (x+y)*c
(x+y)*(b+c) = x*b+y*b + x*c+y*c

This is what the FOIL rule is all about. Though the FOIL rule is limited compared to what the distributive property can do. FOIL only applies to multiplication of two binomials. The distributive property can be applied to multiplying any terms.

The commutative property of multiplication is the idea that you can multiply any two numbers in any order you want
example: 2*3 = 6 and 3*2 = 6
In general: x*y = y*x

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(side note: your formula C = 2πr^2 is incorrect, it should be C = 2πr)

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r = 9

Step-by-step explanation:

First let's find the circumference of Circle A. To use the formula we need radius. Since we have a diameter just divide by 2 to get radius. 27 / 2 = 13.5

Now plug that into C = 2πr: 2π*(13.5) = 27π

The question says that the area of circle B is 3x this value, so Circle B's area must be: 27π * 3 = 81π

Now plug that into A = πr^2 and solve for r:

$81\pi=\pi r^{2}\\81=r^{2}\\\sqrt{81}=r\\9=r$

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Log_(5)(x-4)=1-log_(5)(x-8)

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Answer:

x = 3, x = 9

Step-by-step explanation:

When solving this problem, keep the general format of a logarithm in mind:

$b^x=y\\log_b(y)=x$

Where, (b) represents the base, (x) is the exponent, and (y) is the evalutaor. Please note that others might use slightly different terminotoly than what is used in this answer.

One is given the following expression, and is asked to solve for the parameter (x);

$log_5(x-4)=1-log_5(x-8)$

First, manipulate the exquestion such that all of the logarithmic expressions are on one side. Use inverse operations to do this.

$(log_5(x-4))+(log_5(x-8))=1$

Now use the Logarithmic Base Change rule to simplify. The Logarithmic Base Change rule states the following;

$log_b(x)=\frac{log(x)}{log(b)}$

Remember, if no base is indicated in a logarithm, then the logarithm's base is (10). Apply the Logarithmic Base Change rule to this problem;

$\frac{log(x-4)}{log(5)}+\frac{log(x-8)}{log(5)}=1$

Now remove the denominator. Multiply all terms in the equation by the least common denominator; ( $log(5)$ ) to remove it from the denominator on the left side.

$(\frac{log(x-4)}{log(5)}+\frac{log(x-8)}{log(5)}=1)*(log(5))$

$log(x-4)+log(x-8)=log(5)$

All logarithms have the same base, the left side of the equation has the addition of logarithms. This means that one can apply the Logarithm product rule. The logarithm product rules the following;

$log_b(x*y)=(log_b(x))+(log_b(y))$

This rule can be applied in reverse to simplify the left side of the equation. Rather than rewriting the product of logarithms as two separate logarithms being added, one can rewrite it as one logarithm getting multiplied.

$log(x-4)+log(x-8)=log(5)$

$log((x-4)(x-8))=log(5)$

Now used inverse operations to bring all of the terms onto one side of the equation:

$log((x-4)(x-8))=log(5)$

$log((x-4)(x-8))-log(5)=0$

Similar to the Logarithm product rule, the Logarithm quotient rule states the following;

$log_b(x/y)=(log_b(x))-(log_b(y))$

One can apply this rule in reverse here to simplify the logarithms on the left side:

$log((x-4)(x-8))-log(5)=0$

$log(\frac{(x-4)(x-8)}{5})=0$

The final step in solving this equation is to use the Logarithm of (1) property. This property states the following:

$log_b(1)=0$

When applying this property here, one can conclude that the evaluator must be equal to (1), therefore, the following statements can be made.

$log(\frac{(x-4)(x-8)}{5})=0$

$\frac{(x-4)(x-8)}{5}=1$

Inverse operations,

$\frac{(x-4)(x-8)}{5}=1$

$(x-4)(x-8)=5$

$(x-4)(x-8)-5=0$

Simplify,

$(x-4)(x-8)-5=0$

$x^2-12x+32-5=0$

$x^2-12x+27=0$

Factor, rewrite the quadratic expression as the product of two linear expressions, such that when the linear expressions are multiplied, the result is the quadratic expression:

$x^2-12x+27=0$

$(x-3)(x-9)=0$

Now use the zero product property to solve. The zero product property states that any number times (0) equals (0).

$x=3,x=9$

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