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Kobotan [32]
3 years ago
8

The equation k = 1.6d gives an approximate relationship between d miles and k

Mathematics
1 answer:
uranmaximum [27]3 years ago
6 0

Substituting d= 430 we get,

k = 1.6(430)

k = 688

The distance between Boise, Idaho and Reno, Nevada is 688 kilometers.

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Mrs. Smith had the following transactions during the week:
blsea [12.9K]

Answer:

fghffjfghxhhg

56

Step-by-step explanation:

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8 0
3 years ago
Read 2 more answers
Which is the solution to the inequality?<br> y + 15 less-than 3
Arte-miy333 [17]

Hey there!

"less than" is "<"

y+15<3

Solve it:

y<3-15

y<-12

Hope everything is clear.

Let me know if you have any questions!

Always remember: Knowledge is power!

6 0
1 year ago
Real answers only pls! no links either!
MaRussiya [10]

Answer:

A. 4/8 + 2/4 =1 B.5/8 + 1/4 =0.875

C.6/8 + 3/4 =1.5 D.7/8 + 2/4 =1.375

3 0
2 years ago
Whats the lcd of 1/7,14/17,12/13,5/6
gizmo_the_mogwai [7]
LCD (1/7, 14/7, 12/13, 5/6)

LCM = (7, 7, 13, 6)

=  2 * 3 * 7 * 13

=  546



1/7  = 78/546

14/7  = 1092/546

12/13  = 504/546

5/6  = 455/546




Calculation:

1/7 + 14/7 

= 1 + 14/7

=  15/7


The common denominator you can calculate as the least common multiple of the both denominators:   LCM (7, 7) = 7


Add:

15/7 + 12/13 

= 15 . 13/7. 13  + 12 . 7/13 . 7

= 195/91 + 84/91

=  195 + 84/91 

=  279/91


The common denominator you can calculate as the least common multiple of the both denominators:  LCM (7, 13) = 91



Add:

279/91 + 5/6

= 279 . 6/91 . 6 + 5 . 91/6. 91

= 1674/546 + 455/546

= 1674 + 455/546 

= 2129/546



The common denominator you can calculate as the least common multiple of the both denominators:  LCM (91, 6) = 546



Hence, 546 is the LCM/LCD of (1/7, 14/17, 13/13, 5/6).






Hope that helps!!!!!!


3 0
3 years ago
Solve the following system
scZoUnD [109]

Answer:

{x = -4 , y = 2 ,  z = 1

Step-by-step explanation:

Solve the following system:

{-2 x + y + 2 z = 12 | (equation 1)

2 x - 4 y + z = -15 | (equation 2)

y + 4 z = 6 | (equation 3)

Add equation 1 to equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - 3 y + 3 z = -3 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Divide equation 2 by 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Add equation 2 to equation 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+5 z = 5 | (equation 3)

Divide equation 3 by 5:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 3 from equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y+0 z = -2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{-(2 x) + 0 y+2 z = 10 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 2 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = 8 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = -4 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer:  {x = -4 , y = 2 ,  z = 1

4 0
3 years ago
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