Write a two column proof

Mathematics

1 answer:

NISA [10]3 years ago

7 0

We are given intersecting lines. The measures of a pair of angles forming vertex angles, are 2x+6° and 96°.

The measure of 2 vertex angles is equal, so we solve the equation:

2x+6°=96°, subtract 6° from both sides of the equation
2x=90°, divide both sides of the equation by 2
x=45°

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3 years ago

Read 2 more answers

Multiply (x-3)(4x+2) using the distributive property.

asambeis [7]

$\text{The distributive property:}\ (a+b)(c+d)=ac+ad+bc+bd\\\\(x-3)(4x+2)=(x)(4x)+(x)(2)-(3)(4x)-(3)(2)\\\\=4x^2+2x-12x-6=4x^2-10x-6$

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3 years ago

Which set of numbers is arranged in order from least to greatest?

charle [14.2K]

The answer is: [C]: -0.7, ⅕, 0.35, ⅔ .
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Explanation:
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Note that in this correct Answer choice "C" given, we have the following arrangement of numbers:
_____________________________________________________
→ -0.7, ⅕, 0.35, ⅔ ;
______________________________________
We are asked to find the "Answer choice" (or, perhaps, "Answer choices?") given that show a set of numbers arranged in order from "least to greatest"; that is, starting with a value that is the smallest number in the arrangement, and sequentially progressing, in order from least to greatest, with the largest (greatest) number in the arrangement appearing as the last number in the arrangement.
______________________
Note the EACH of the 4 (four) answer choices given consists of an arrangement with ONLY one negative number, "- 0.7". Only TWO of the answer choices—Choices "B" and "C"—have an arrangement beginning with the number, "-0.7 "; So we can "rule out" the "Answer choices: [A] and [D]".
________________________
Let us examine: Answer choice: [B]: -0.7, 0.35, ⅕, ⅔ ;
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Note: The fraction, "⅕" = "2/10"; or, write as: 0.2 .
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The fraction, "⅔" = 0.6666667 (that is 0.6666... repeating; so we often see a "final decimal point" rounded to "7" at some point.
___________________________________________
Through experience, one will be able to automatically look at these 2 (two) fractions and immediately know their "decimal equivalents".
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Otherwise, one can determine the "decimal form" of these values on a calculator by division:
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→ ⅕ = 1/5 = 1 ÷ 5 = 0.2
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→ ⅔ = 2/3 = 2 ÷ 3 = 0.6666666666666667
___________________________________
For Answer choice: [B], we have:
______________________________
→ -0.7, 0.35, ⅕, ⅔ ;
_________________________
→ So, we can "rewrite" the arrangement of "Answer choice [B]" as:
___________________________________________
→ -0.7, 0.35, 0.2, 0.666666667 ;
________________________________
→ And we can see that "Answer choice: [B]" is INCORRECT; because
"0.2" (that is, "⅕"), is LESS THAN "0.35". So, "0.35" should not come BEFORE "⅕" in the arrangement that applies correctly to the problem.
_______________________________________
Let us examine: Answer choice: [C]: -0.7, ⅕, 0.35, 0.666666667 .
____________________________________________
→ Remember from our previous— and aforementioned—examination of "Answer Choice: [B]" ; that:
____________________________
→ ⅕ = 0.2 ; and:
→ ⅔ = 0.666666667
_______________________
So, given:
____________
→ Answer choice: [C]: -0.7, ⅕, 0.35, ⅔ ;
______________________
→ We can "rewrite" this given "arrangement", substituting our known "decimal values for the fractions:
______________________________
→ Answer choice: [C]: -0.7, 0.2, 0.35, 0.666666667 ;
_________________________________________
→ As mentioned above, this sequence starts with "-0.7", which is the ONLY negative number in the sequence; as such, the next positive number is correct. Nonetheless, "0.2" (or, "(⅕") is the next number in the sequence, and is greater than "-0.7". The next number is "0.35. "0.35" is greater than "⅕" (or, "0.2"). Then next number is "(⅔)" (or, "0.666666667").
"(⅔)"; (or, "0.666666667") is greater than 0.35.
____________________________
This set of numbers: "-0.7, ⅕, 0.35, ⅔" ; is arranged in order from least to greatest; which is "Answer choice: [C]: -0.7, ⅕, 0.35, ⅔" ; the correct answer.
________________________________________________________

6 0

3 years ago

B to the eighth power multiplied by b to the fourth power

scZoUnD [109]

U just add the exponents.

b^8 x b^4 = b^12

hope this helps :)

7 0

3 years ago

Need help please help me

ANTONII [103]

Megan:
x to the one third power = $x ^{1/3}$
x to the one twelfth power = $x ^{1/12}$

The quantity of x to the one third power, over x to the one twelfth power is:
 $\frac{x ^{1/3}}{x ^{1/12}}$

Since $\frac{ x^{a} }{ x^{b} } = x ^{a-b}$
then $\frac{x ^{1/3}}{x ^{1/12}} = x^{1/3-1/12}$

Now, just subtract exponents:
1/3 - 1/12 = 4/12 - 1/12 = 3/12 = 1/4

$\frac{x ^{1/3}}{x ^{1/12}} = x^{1/3-1/12} = x^{1/4}$

Julie:
x times x to the second times x to the fifth = x * x² * x⁵

The thirty second root of the quantity of x times x to the second times x to the fifth is
 $\sqrt[32]{x* x^{2} * x^{5} }$

Since $x^{a}* x^{b}= x^{a+b}$
Then $\sqrt[32]{x* x^{2} * x^{5} }= \sqrt[32]{ x^{1+2+5} } =\sqrt[32]{ x^{8} }$

Since $\sqrt[n]{x^{m}} = x^{m/n} }$
Then $\sqrt[32]{ x^{8} }= x^{8/32} = x^{1/4}$

Since both Megan and Julie got the same result, it can be concluded that their expressions are equivalent.

3 0

3 years ago