Answer:
20
Step-by-step explanation:
y+4=−12(x−2)
Step 1: Add -4 to both sides.
y+4+−4=−12x+24+−4
y=−12x+20
Answer:
195.6 mph, 069 degrees
Step-by-step explanation:
two different ways, same result
We have that
<span>Circle 1: center (8, 5) and radius 6
</span><span>Circle 2: center (−2, 1) and radius 2
we know that
the equation of a circle is
(x-h)</span>²+(y-k)²=r²
for the circle 1---------> (x-8)²+(y-5)²=36
for the circle 2---------> (x+2)²+(y-1)²=4
using a graph tool
see the attached figure
Part A)<span>What transformations can be applied to Circle 1 to prove that the circles are similar?
we know that
r1/r2---------> 6/2------> 3
</span><span>
to prove that the circle 1 and circle 2 are similar, the radius of circle 1 </span>must be divided by 3 and translate the center of the circle 1 (10) units to the left and (4) units down
<span>
the answer part A) is
</span>
the radius of circle 1 must be divided by 3 and translate the center of the circle 1 (10) units to the left and (4) units down
Part B) <span>What scale factor does the dilation from Circle 1 to Circle 2 have?
the answer Part B) is
the scale factor is (3/1)</span>
Answer:
n squared + 3n + 1
Step-by-step explanation:
5,11,19,29
Firstly look at the difference between each number. The first difference is 6 then 8 then 10 etc. After that you look at your created sequence - 6,8,10 etc. The difference is 2 each time. Then applying rules, you have to do the constant difference divided by 2 to get a coefficient of n squared. So in this case it's n squared because 2/2 = 1 so you don't have to place a 1 in front of the n squared. After you create a sequence from the n squared. That would be 1,4,9 etc. Then you need to see how to get from the sequence: 1,4,9 etc to your original sequence: 5,11,19 etc. So if you calculate it you will get 4,7,10 because firstly 5-1 = 4 then 11-4 = 7 etc. The sequence 4,7,10 is a linear sequence so the constant difference is 3 each time. So to get a nth term of a linear sequence you will start off as 3n then you will substitute 1 then 2 then 3 into the 3n. Therefore that would be 3,6 etc. So if you take the first substituted term, that would be 3 as said before then you will have to see how to get from the 3 to 4 so that is just adding 1. So the nth term of this linear sequence is 3n + 1. Check if it works at the end. So the overall nth term of the quadratic sequence is n squared as said before + 3n + 1.
To show equality with fractions both fractions should be converted to show their values with the same denominator then draw a picture of those fractions.
Remember that you can only compare fractions with the same denominator as is said with other operations dealing with fractions.