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love history [14]
3 years ago
6

Help pls!! I only have 1/2 hours to finish :(( on the scale drawing of a car, the length from bumper to bumper is 6 inches. the

car’s actual length, from bumper to bumper, is 16 feet. using I to represent inches and to F represent feet, write two equations to express the rate relationship
Mathematics
1 answer:
notka56 [123]3 years ago
3 0

Answer:

I = 3\8 . F and F = 8\3 . I

Step-by-step explanation:

Hope this helps! ;3

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1x + 2 - 3 = 8 <br><br><br> what is x?
valina [46]

Answer:

x= Multiplication

Step-by-step explanation:

1 x +2 -3 =8

8 0
3 years ago
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a house on the market was valued at 446000 after several years of value increased by 15% by the how much did the house value inc
olga_2 [115]

Answer:

the house will have gained $66,900$ worth of value putting the house worth at $512,900$

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446,000  ➗  100 X  15 = 66,900

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hope this help if you need any more help P.M. me.

6 0
3 years ago
11 + 5 + 22 × 12 − 10
Masteriza [31]
Follow PEMDAS

22 x 12 = 264

11 + 5 = 16

264 + 16 = 280

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270 is your answer

hope this helps
6 0
4 years ago
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Write each of the following expression without using absolute value. <br> |6m-54|, if m&lt;9
Simora [160]

9514 1404 393

Answer:

  54 -6m

Step-by-step explanation:

For values of m < 9, the argument of the absolute value function is less than 0. Then the expression evaluates to ...

  -(6m -54) = 54 -6m

3 0
3 years ago
A "Local" train leaves a station and runs at an average rate of 35 mph. An hour and a half later an "Express" train leaves the s
pshichka [43]
Recall your d = rt, distance = rate * time.

so...Local say "L" is going at a speed of 35mph...ok... and Express or "X" is going at 56mph.

by the time the two trains meet, and X is ready to overtake L, the distance that both have travelled, since is a parallel road, is the same, say "d".  So if L has travelled "d" miles, then X had travelled "d" miles too, over the same road, maybe different lane.

now, because X left 1 1/2 hour later, by the time they meet, say X has been running for "t" hours, but because it left 1 1/2 hour later, L has been running for " t + 1 1/2 " hours, or " t + 3/2 " hours.

\bf \begin{array}{lccclll}&#10;&distance&rate&time\\&#10;&-----&-----&-----\\&#10;Local&d&35&t+\frac{3}{2}\\&#10;Express&d&56&t&#10;\end{array}&#10;\\\\\\&#10;\begin{cases}&#10;\boxed{d}=35\left( t+\frac{3}{2} \right)\\&#10;d=56t\\&#10;----------\\&#10;\boxed{35\left( t+\frac{3}{2} \right)}=56t&#10;\end{cases}&#10;\\\\\\&#10;35t+\cfrac{105}{2}=56t\implies \cfrac{105}{2}=21t\implies \cfrac{105}{42}=t&#10;\\\\\\&#10;\cfrac{5}{2}=t\implies 2\frac{1}{2}=t

so, they met 2 and a half hours later after X left, and a milllisecond later X overtook L.
4 0
3 years ago
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