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Sergio039 [100]
3 years ago
12

How many multiples of 5 are there between 199 and 1,198? Hint: an = a1 + d(n − 1), where a1 is the first term and d is the commo

n difference.
Mathematics
1 answer:
GaryK [48]3 years ago
4 0
An=a1+d (n-1)
A1=200 since that's the first term that can be a multiple of 5
N=? That's what we need to find
An=1195 since that's the last multiple of 5 that we can use
D=5
Plug in
1195=200+5 (n-1)
-200 both sides
995=5 (n-1)
Distribute 5 to n-1
995=5n-5
+5 both sides
1000=5n
÷5 both sides
N=200 there are 200 multiples of 5 in between 199 and 1198
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Answer:

Step-by-step explanation:

Let's write a system of equations. I'm going to call them w, x, y, and z.

w = 4x (because m∠1 is 4 times m∠2)

w + x = 90 (bc ∠1 and ∠2 are complementary, meaning they add up to 90)

w = y (bc ∠1 and ∠3 are vertical angles, meaning they are equal)

y + z = 180 (bc ∠3 and ∠4 are supplementary, meaning they add up to 180)

When we substitute w=4x into the second equation, we get 5x=90 → m∠2 = 18°. Since w=4x, m∠1 = 18*4 = 72°. Since w=y, m∠3 = 18°. Finally, because y+z=180, m∠4 = 180-18 = 162°. Hope this helps!

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3 years ago
Solve: -2/5 (1/4 x -4)
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2/5 is tje correct answer for your question

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Step-by-step explanation:

Hey There!

for this question they would like you to simplify the expression using distributive property

To do this the first step is to distribute the 9 in to everything that is in the parenthesis

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the next step is to combine like terms

the only like terms in the expression are 45x and 4x so we just add the two together

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