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Natalija [7]
3 years ago
10

What would '6 meters to 60 meters' be in a ratio

Mathematics
1 answer:
Anna007 [38]3 years ago
3 0
Because you van turn 6 into 60 so they are in common. Hope that helps

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Find the product -a2 b2 c2 (a+b-c)
mina [271]
Use the distributive property:
a(c+b)=ab+ac
And
a^n\cdot a^m=a^{n+m}\\\\a=a^1

-a^2b^2c^2(a+b-c)=-a^2ab^2c^2-a^2b^2bc^2+a^2b^2c^2c=-a^3b^2c^2-a^2b^3c^2+a^2b^2c^3

7 0
3 years ago
What are the factors of -9 that have a sum of 0
Bas_tet [7]

Answer:

3 and -3

Step-by-step explanation:

+3 -3 =0

sum mean adding

7 0
3 years ago
The stem-and-leaf plot shows the scores of a class of sixth graders on a math test.
SOVA2 [1]

A stem and leaf plot shows sets of two digit numbers, by separating the ten’s place and the one’s place. On the left is the different ten’s values, while on the right next to each of the values on the left is the one’s values that associate with each of the ten’s values. This means that the numbers in this set of data are 32, 47, 51, 55, 55, 55, 58, 64, and so on. From there, you can use that knowledge to figure out how many scores were above 60.


The terms that are above 60 are 64, 65, 73, 74, 77, 87, 88, 91, 93, 93, 97, 99, and 99, for a total of 13 of the 20 scores being above 60.

8 0
3 years ago
Gina and Rhonda work for different real estate agencies. Gina earns a monthly salary of $5,000 plus a 6% commission on her sales
KATRIN_1 [288]
The answer is
B $75,000
3 0
4 years ago
Tan 3A in terms of tan​
Zanzabum

Here's the sum rule for the tangent function:

\tan(a+b)=\dfrac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}

In the special case a=b, this becomes the double angle formula:

\tan(a+a)=\tan(2a)=\dfrac{\tan(a)+\tan(a)}{1-\tan(a)\tan(a)}=\dfrac{2\tan(a)}{1-\tan^2(a)}

In your case, you case use the sum rule once:

\tan(3a)=\tan(2a+a)=\dfrac{\tan(2a)+\tan(a)}{1-\tan(2a)\tan(a)}

And use it again, in the special case of the double angle:

\dfrac{\dfrac{2\tan(a)}{1-\tan^2(a)}+\tan(a)}{1-\dfrac{2\tan(a)}{1-\tan^2(a)}\tan(a)}

We can obvisouly simplify this expression a lot: let's deal with the numerator and denominator separately: the numerator is

\dfrac{2\tan(a)}{1-\tan^2(a)}+\tan(a) = \dfrac{2\tan(a)+\tan(a)-\tan^3(a)}{1-\tan^2(a)}

and the denominator is

1-\dfrac{2\tan(a)}{1-\tan^2(a)}\tan(a) = \dfrac{1-\tan^2(a)-2\tan^2(a)}{1-\tan^2(a)} = \dfrac{1-3\tan^2(a)}{1-\tan^2(a)}

So, the fraction is

\dfrac{2\tan(a)+\tan(a)-\tan^3(a)}{1-\tan^2(a)}\cdot \dfrac{1-\tan^2(a)}{1-3\tan^2(a)} = \dfrac{2\tan(a)+\tan(a)-\tan^3(a)}{1-3\tan^2(a)}

8 0
3 years ago
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