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Kipish [7]
3 years ago
6

Anne is painting her house light blue. To make the color she wants, she must add 3 cans of white paint to every 2 cans of blue p

aint. How many cans of white paint will she need to mix with 6 cans of blue?
Mathematics
1 answer:
GuDViN [60]3 years ago
6 0

Answer:

9 cans of white paint

Step-by-step explanation:

To solve this you can set up a ratio.

The ratio right now is 3 : 2, you have 3 cans of white paint for every 2 cans of blue paint. This question is asking how many white paint cans would be needed for 6 cans of blue paint. You can see the relationship between the number of blue paint cans and the new number of blue paint cans, maybe it's multiplying by 4 or 2 for example, and once we find that out we can do the same exact thing to the white cans.

\frac{3}{2}  = \frac{?}{6}

We can see that to get from 2 to 6 you multiply by 3, so now we do that to the other side of the fraction as well, we multiply by 3. 3 multiplied by 3 is 9, so if we were to have 6 cans of blue paint we would need 9 cans of white paint to get that perfect shade of light blue. Anne would need 9 cans of white paint if she had 6 cans of blue paint to make her shade of blue.

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3 years ago
The first and second terms of a linear sequence (A.p) are 3 and 8 respectively. Please determine the least Number of terms of Ap
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Answer:

To rewrite your question, you are looking for the smallest    that fulfills the inequality:

∑=1>250

First, we must find the sequence    in explicit form. We can think of it as a line we are trying to get the slope-intercept form of. We have the points (1, 3) and (2, 8). Therefore, the slope is  8−32−1=51=5 . Now we just need the y-intercept. We can find it through substitution:

=5+

In order to solve for   , we can plug in one of the points that we were already given. I will choose the point (1, 3). Note that this point just means that when  =1 ,  =3 .

3=5(1)+

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Now for the original question.

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2+15+1100>100+1100

(+110)2>100+1100

(+110)2>10001100

+110>±10001100‾‾‾‾‾‾‾√

+110>±11010001‾‾‾‾‾‾√

>−110±11010001‾‾‾‾‾‾√

Since we are looking for the smallest value, we will subtract for the plus or minus sign.

>−110−11010001‾‾‾‾‾‾√≈−10

Since that is negative, we will have to add instead.

>−110+11010001‾‾‾‾‾‾√≈9.9

Therefore, the smallest integer    that makes sense and satisfies the inequality is 10.

So, the first 10 numbers must be added.

Honestly, it would have been quicker just to brute-force this.

Step-by-step explanation:

Have a good day

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