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Wewaii [24]
3 years ago
6

A 1.51 gram bar of gold is worth 1567.01 dollars. what is the value in dollar of a 2.23 gram of gold ingot?

Mathematics
1 answer:
marin [14]3 years ago
7 0

Answer:

Step-by-step explanation:

1.51 gram to $ 1567.01 = 2.23 gram to $ x

1.51 / 1567.01 = 2.23 / x

cross multiply because this is a proportion

(1.51)(x) = (1567.01)(2.23)

1.51x = 3494.43

x = 3494.43 / 1.51

x = $ 2314.19 <===== what 2.23 grams of gold is worth

==========================

x = first car and y = new car

x = 28,033.51

y = 2x - 301.26

y = 2(28,033.51) - 301.26

y = 56,067.02 - 301.26

y = $ 55,765.76 <====== new car cost

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15 multiplied by 11 is 165. so you're width is 11.
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Someone help me please
pogonyaev

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10x

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8 0
3 years ago
Read 2 more answers
Jan's age is 3 years less than twice Tritt's age. The sum of their ages is 30. find their ages
Blizzard [7]
Let Tritt's age = t & Jan's age = 2t - 3

if the sum of their ages equals 30 then..
t + 2t - 3 = 30
combine terms and simplify
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3 0
3 years ago
Help please!!! Thank you
ANTONII [103]

Answer:

Option (G)

Step-by-step explanation:

Let the length of the race = a miles

Since, Speed = \frac{\text{Distance}}{\text{Time}}

Time taken to cover 'a' miles with the speed = 12 mph,

Time taken 't_1' = \frac{a}{12}

Time taken to cover 'a' miles with the speed = 11 mph,

Time taken 't_2' = \frac{a}{11}

Since the time taken by David to cover 'a' miles was 10 minutes Or \frac{1}{6} hours more than the time he expected.

So, t_2=t_1+\frac{1}{6}

\frac{a}{11}=\frac{a}{12}+\frac{1}{6}

\frac{a}{11}-\frac{a}{12}=\frac{1}{6}

\frac{12a-11a}{132}=\frac{1}{6}

a = 22 mi

Therefore, distance of the race = 22 mi

Option (G) is the correct option.

6 0
3 years ago
2x+11y=22 <br><br> -2x-11y=50
Harman [31]

Answer:

Step-by-step explanation:

2x + 11y = 22

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0 ≠ 72

no solution

7 0
3 years ago
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