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Ray Of Light [21]
3 years ago
5

(10x-20) (6x+8) what is x?

Mathematics
1 answer:
Lilit [14]3 years ago
7 0
X= 2 and x= -8/6 if you make it = 0 
(10x-20)(6x+8)=0
10x-20=0
10x=20
x=2
6x+8=0
6x=-8
x=-8/6
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3 years ago
If the perpendicular bisector of one side of a triangle goes through the opposite vertex, then is the triangle isosceles?
Ne4ueva [31]

Answer:

True

Step-by-step explanation:

The perpendicular bisector of the opposite side to the vertex bisects the angle at the vertex into two equal parts and also bisects the triangle into two equal parts.

Let A be the angle at the vertex, then assume that the angle is an isosceles triangle with base angles B.

We need to show that A = 180 - 2B for an isoceles triangle

The perpendicular bisector bisects A into two so the new angle in the vertex one half of the bisected triangle is A/2.

Since this half triangle is a right-angled triangle, the third angle in it is 90.

So, A/2 + B +  90 = 180 (Sum of angles in a triangle)

subtracting 90 from both sides, we have

A/2 + B + 90 - 90 = 180 - 90

A/2 + B = 90

subtracting B from both sides, we have

A/2 + B = 90

A/2 = 90 - B

multiplying through by 2, we have

A = 2(90 - B)

A = 180 - 2B

Since A = 180 - 2B, then our triangle is an isosceles triangle.

7 0
3 years ago
Dale deposits $4000 into an account that pays simple interest at a rate of 6% per year. How much interest will he be paid in the
BartSMP [9]

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$1200

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3 years ago
Please answer ASAP. The question is down below
Nadusha1986 [10]

Answer:  1) D      2) B

<u>Step-by-step explanation:</u>

1) The denominator cannot equal zero, Factor the denominator to find the zeros.  n³ - 4n² + 3n = 0

           n(n - 3)(n - 1) = 0

       n=0  n-3=0   n-1=0

       n=0    n=3      n=1

2) In order to be a rational expression, it must be in the form \dfrac{a}{b} where both a and b are integers.

Since \sqrtb\sqrt b is irrational and not an integer, then the expressions containing an irrational term <em>after simplified</em> cannot result in an integer.

Therefore, options <em>(i)</em> and <em>(iv)</em> are not rational expressions.

Option <em>(iii)</em> contains b as an exponent.  Since there is no information about b, it could be a fraction, which means it could be an irrational number.

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3 years ago
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