what is 7 over (fraction bar) (8×4)+90÷9 ÷ (6^3)^-2 × 9 over (fraction bar) 6(5+4) = ?????
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Note: " (8 * 4) + 90 ÷ 9 " = " (8 * 4) + (90 ÷ 9) " = "32 + 10 = 42" .
We can rewrite the "fraction":
7 / [ (8 * 4) + 90 ÷ 9 ] ; as: [tex] \frac{7}{42}[/tex] ;
and can reduce this value to (and rewrite as): " [tex] \frac{1}{6}[/tex] " ;
{ since: " [tex] \frac{7}{42}[/tex] = [tex] \frac{7/7}{42/7} [/tex] = [tex] \frac{1}{6}[/tex]" .}.
Note: The "fraction" : [tex] \frac{9}{6(5+4)} [/tex] ; the "9/6" cancel to "3/2" ;
{Since: "9÷3 = 3" ; and since: "6÷3 = 2" ).
So rewrite: [tex] \frac{9}{6(5+4)} [/tex] ; as:
[tex] \frac{3}{2(5+4)} [/tex] ;
Note: [tex] \frac{3}{2(5+4)} [/tex] ;
= [tex] \frac{3}{2(9)} [/tex] ;
= [tex] \frac{3}{18} [/tex] ;
→ Note: [tex] \frac{3}{18} [/tex] ; can be reduced to; & rewritten as:
" [tex] \frac{1}{6} [/tex] " ;
→ {since: " [tex] \frac{3}{18} [/tex] = [tex] \frac{3/3}{18/3} [/tex] ;
= [tex] \frac{1}{6} [/tex] ;
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Note that: (6²) ⁻² = 6⁽ ² * ⁻²⁾ = 6⁻ ⁴ ;
= [tex] \frac{1}{6^4} [/tex] ;
_________________________________________________________
So, we can rewrite our expression as:
[tex] \frac{1}{6}[/tex] ÷ [tex] \frac{1}{6^4} [/tex] * [tex] \frac{1}{6}[/tex] ;
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Note:
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[tex] \frac{1}{6}[/tex] ÷ [tex] \frac{1}{6^4} [/tex] * [tex] \frac{1}{6}[/tex] ;
= ( [tex] \frac{1}{6}[/tex] ÷ [tex] \frac{1}{6^4} [/tex] ) * [tex] \frac{1}{6}[/tex] ;
= ( [tex] \frac{1}{6}[/tex] * [tex] \frac{6^4}{1} [/tex] ) * [tex] \frac{1}{6}[/tex] ;
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Start with :
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→ ( [tex] \frac{1}{6}[/tex] * [tex] \frac{6^4}{1} [/tex] ) ;
The "6" in the " [tex] \frac{1}{6}[/tex] " ; cancels to a "1" ;
and the "6⁴ " in the " [tex] \frac{6^4}{1} [/tex] ) " ; cancels to a "6³ " ;
→ {Since: "6⁴ / 6 = 6⁴ / 6¹ = 6⁽⁴ ⁻ ¹⁾ = 6³ ;
→ and we have: [tex] \frac{1}{1}[/tex] * [tex] \frac{6^3}{1} [/tex] ;
→ The " [tex] \frac{1}{1}[/tex] " can be eliminated; since "1÷1= 1 " ;
and since: any value, divided by "1", equal that same value; & since "any value, multiplied by "1"; equals that same value" ;
________________________________________________________
We have: " 6³ " .
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Now, we multiplied this value, " 6³ " , by " [tex] \frac{1}{6}[/tex] " ;
→ 6³ * [tex] \frac{1}{6}[/tex] ;
= [tex] \frac{6^3}{1} [/tex] * [tex] \frac{1}{6}[/tex] ;
Note: The "6³ " cancels out to "6² " ; and the "6" cancels out to "1" ;
{since: "6³ ÷ 6 = 6³ ÷ 6¹ = 6⁽³ ⁻ ¹⁾ = 6² " ;
and since: "6 ÷ 6 = 1 " ; ________________________________________________________
→ The answer is: 6² = 6 * 6 = 36 ;
________________________________________________________
The answer is: " 36 " .
________________________________________________________
Related Questions
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Answer: Comparison bars are a display tool for solving the scenario in which two elements are compared with the difference is either known or unknown.
There are three kinds of use of comparison bars in word problem such as 1) If difference is unknown then A-B=Difference[where A -bigger quantity, B-smaller quantity]. 2) If bigger quantity is unknown then A=Difference-B 3) If smaller quantity is unknown then B=Difference-A.
A drawing of comparison bars, also known as a bar graph, can be used in word problems when there is a need to compare different entities or categories. It is particularly useful in situations where quantitative data needs to be visually represented and compared. For example, if you are given data about the sales of different types of fruits in a grocery store, you can use a bar graph to compare the sales of each fruit.
Explanation:A drawing of comparison bars, also known as a bar graph, can be used in word problems when there is a need to compare different entities or categories. It is particularly useful in situations where quantitative data needs to be visually represented and compared. For example, if you are given data about the sales of different types of fruits in a grocery store, you can use a bar graph to compare the sales of each fruit.
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he graph of the function f(x) = –(x + 6)(x + 2) is shown below.
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Answer:
Which statement about the function is true?
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a group of 4 students are selling candy bars to raise money for a field trip. the group needs to sell 53 candy bars. if each student sells an equal amount, how many candy bars do they get in all? interpret your remainder
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Answer:
Each student is gonna sell 13 candy bars, and there is gonna be 1 bar unsold, because they sold equal amount of candy bars.Step-by-step explanation:
If there are 4 students, they need to sell 53 candy bars, and they will sell the same amount each one, we just have to divide:
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Because the division is not exact, we have a remainder of 1.
This can be interpreted as 1 candy is not sold, because they are gonna sell equally, so in that case each one would sell 13 candy bars, and one will be unsold.
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Answers
Final answer:
Stephanie is right; when you multiply decimals less than one, the result is smaller than the original numbers. The multiplication of (-0.04)(-0.1) is 0.004, (0.03)(-0.02) is -0.0006, and (0.7)(0.4) is 0.28. These examples illustrate the rule correctly.
Explanation:
Stephanie is correct in her explanation to Stacey. When multiplying two numbers where one or both numbers are less than one, the product will be smaller than the original numbers. This is because you are essentially taking a fraction of each number. Let's confirm this by calculating the products of the given examples.
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The number that would have 12. 75 being 25 percent of it would be 51.
How to find the number ?To find the number of which 12. 75 is 25 %, you can set up a proportion and solve for the unknown number ( let's call it " x ").
You can write the proportion as:
( 25 / 100 ) * x = 12. 75
Now, simplify the left side of the equation:
( 1 / 4 ) * x = 12.75
To solve for x, multiply both sides by 4 to isolate x:
x = 12. 75 * 4
x = 51
So, 12. 75 is 25% of the number 51.
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w=width
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#2- The equation is 4 + 5x = 1. To get "5x" by itself on that side, let's subtract 4 to that side and the right side. After you do that, it'll be: 5x = -3. Then divide "x" by 5 to each side. The answer is: x= -3/5 (-0.6).
#3- The equation is 4y - 15 = -10. If we want "4y" by itself, we will do the opposite of subtracting 15 to adding 15 to both sides. After you do that, it'll be: 4y = 5. To finish it off, divide "y" by 4 to both sides. That will get you with:
y = 5/4 (1.25).
#4- The equation is 3 - 4d = 13. To get 4d by itself, let's subtract 3 to each side. That'll leave you with: -4d = 10. Now divide d by "-4" to both sides. The answer is: d = -10/4 (-2.5).
#5- The equation is 8 = 9x - 7. Let's add 7 to each side so that 9x can be by itself. After we do that, it's simplified to: 15 = 9x. Finish it off by dividing "x" by 9 to both sides. That'll leave you with the answer: x = 15/9 (in decimal form, this decimal would be repeating {1.66666}).
Anyone can fix my mistakes if they are wrong. I hope this helps!
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An architect is designing a house in which a 104 square foot room and a 130 square foot room share a wall what is the greatest possible integer length of the shared wall -- THIS QUESTION WAS ASKED AND ANSWERED. BUT ANSWERED INCORRECTLY. NOTE THE ANSWER GIVEN CHANGED THE DIMENSIONS TO 140 FROM 104. AND THIS WEBSITE WILL NOT ALLOW ME TO REPORT THAT.
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Then, the area of both squares have x as common factor.
So, x is the greatest common factor of 104 and 130.
You should know how to calculate the greatest common factor of two integers.
Just find the prime factors and choose the common factors raised to the lowest exponent.
104 = (2^3) (13)
130 = (2) (5)(13)
=> the greatest common factor is 2 * 13 = 26, and that is the greatest possible integer length of the shared wall.
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I hope this helps!
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