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diamong [38]
3 years ago
5

Help please! P - 1/4 * 3 = -5/6 What is the answer??

Mathematics
2 answers:
Sloan [31]3 years ago
6 0
The answer is -1/12. I did this exact question and I got it right.
Good luck though!
Korvikt [17]3 years ago
3 0
I think answer is -1/6............
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Lee's family had a garden in the shape of a right triangle. the total area of the garden is 16 square feet. if they expand the g
lana [24]
Area of right-angled triangle  is given by;

Area, A = 1/2 *b*h, Where b=base, h=height

Therefore,

A1 = 1/2bh = 16 in^2
A2 = 1/2 (2b)(2h) = 2bh

Ratio of increase = A2/A1 = {2bh}/(1/2bh} = 4 (the area is increased 4 times)
The,
A2 = 4*16 = 64 in^2
Therefore,
The area is increased by (64-16) = 48 in^2
5 0
3 years ago
I need help solving unit rates
guapka [62]
It takes Selma 1/6 of a hour to run 3/4 miles.
Just multiply 3/4 by 6 to get 18/4 which is 4.5 miles or 4 1/2 mile
7 0
3 years ago
A school is preparing fruit baskets for a local nursing home. There are 162 apples, 108 oranges, and 180 bananas. If the baskets
kari74 [83]

To solve this exercise, we must calculate the Maximum Common Divisor (M.C.D.), which is the greatest common divisor of 62,108 and 180. We apply factor decomposition, and we obtain:

 162= (3^4)(2)

 108= (2^2)(3^3)

 180= (2^2)(3^2)(5)

 We choose the common numbers with their lowest exponent:

 M.C.D.= (2)(3^2)

 M.C.D.= 2x9

 M.C.D.= 18

 The greatest number of baskets that can be made is: 18 baskets

 To know how many apples, oranges and bananas are in each basket, we must divide the greatest number of them that can be made between the total amount of apples, oranges and bananas, as shown below:

 162/18= 9 apples

 108/18= 6 oranges

 180/18= 10 bananas

 So, there are 9 apples, 6 oranges and 10 bananas in each basket.


3 0
4 years ago
Read 2 more answers
If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

1052-864

188

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, 6^{2}=36 and 36\times 5=180, 36\times 6=216. This implies that 36\times 6 exceeds 188 and thus there can be at most 5 groups of 6^{2}.

Then,

188-5\times6^{2}

188-5\times36

188-180

8

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, 6^{1}=6 and 6\times 1=6, 6\times 2=12. This implies that 6\times 2 exceeds 8 and thus there can be at most 1 group of 6^{1}.

Then,

8-1\times6^{1}

8-1\times6

8-6

2

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

Then,

1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0
3 years ago
It is not always helpful to rewrite an equation in standard form as a first step in solving it.
Kitty [74]
The answer to the problem above is True. Standard form of an equation means the right side of the equation should be 0, and everything on the left side. For example, x^2 = 4 , to make this this in standard form you have to transfer number 4 to the left side of the equation making it x^2 - 4 = 0.
In this situation, if you will be asked to look for x, there is no need to make the equation in standard form, you can directly take the square root of both sides of the equation to get x.
7 0
3 years ago
Read 2 more answers
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