Answer:
I shall assist thee.
Step-by-step explanation:
Being me, I would say the answer is A. or B. But i'd prolly go with A tbh.
And oofers on me if this ain't what you need.
The answer is 30.
You need to use sine here
So:
You can set up a proportion to determine the length of the enlarged photo
Hi there!
The correct answer is 6.25 times.
This is how you get there:
1) First off, you are going to need to have the same unit (either cm or mm). I used millimeters for this problem. 25cm=250mm
2) This next part is very easy to get wrong. You need to figure out what you are dividing, and by what. Simply look at the problem. <span><em>"How much longer is the guinea pig than her fish?" </em>This shows what numbers you should be getting. The guinea pig is 250mm long, where the fish is only 40. This means that her guinea pig should be larger. 250÷40=6.25
If you did this problem the other way, you would've ended up with .16. If you put the sizes into perspective, you will notice that this is highly unlikely.
I hope that I have been able to simplify this problem for you!
Brady
</span>
<span>(3x - 1)( x + 5)(4x - 3) = 12x^3 + 47x^2 - 62x +15 </span>
<span>(3x^2 –
x + 15x - 5)( 4x – 3 ) = <span>12x^3 + 47x^2 - 62x +15 </span></span>
<span>(3x^2 +
14x - 5)( 4x – 3 ) = <span>12x^3 + 47x^2 - 62x +15 </span></span>
<span>12x^3 –
9x^2 + 56x^2 – 42x – 20x + 15 = <span>12x^3 +
47x^2 - 62x +15 </span></span>
<span>12x^3 –
47x^2 - 62x + 15 = <span>12x^3 + 47x^2 - 62x +15 </span></span>
<span> </span>
