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Mkey [24]
2 years ago
10

What is the value of 0.6 - (0.7)(1.4)

Mathematics
1 answer:
sineoko [7]2 years ago
7 0

Answer:

-0.38

Step-by-step explanation:

0.6-(0.7)(1.4)

0.6-0.98

-0.38

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Zarrin [17]

Answer:

6

Step-by-step explanation:

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3 years ago
Betsy and 3 of her friends are splitting a whole watermelon. There are 6 circular slices of watermelon. How many slices of water
ycow [4]
So you are going to do 3 divided by 6
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I am a number greater than 40,000 and less than 60,000. My ones digit and tens digit are the same. My ten-thousands digit is 1 l
jarptica [38.1K]
I am a number greater than 40,000 and less than 60,000:

40,000 < n < 60,000

This means that:

n = 10,000n₁ + 1,000n₂ + 100n₃ + 11n₄

And also:

4 ≤ n₁ < 6

0 ≤ n₂ ≤ 9

0 ≤ n₃ ≤ 9

0 ≤ n₄ ≤ 9

My ten thousands digit is 1 less than 3 times the sum of my ones digit and tens digit:

n₁ = 3*2n₄ - 1

n₁ = 6n₄ - 1

This means that:

n = 10,000*(6n₄-1) + 1,000n₂ + 100n₃ + 11n₄

n = 60,000n₄ - 10,000 + 1,000n₂ + 100n₃ + 11n₄

n = 60,011n₄ - 10,000 + 1,000n₂ + 100n₃

<span>My thousands digit is half my hundreds digit, and the sum of those two digits is 9:

n</span>₂ = 1/2 * n₃
<span>
n</span>₂ + n₃ = 9
<span>
Therefore:

n</span>₂ = 9 - n₃
<span>
Therefore:

9 - n</span>₃ = 1/2 * n₃
<span>
9 = 1/2 * n</span>₃ + n₃
<span>
9 = 1.5 * n</span>₃
<span>
Therefore:

n</span>₃ = 6
<span>
If n</span>₃=6, n₂=3.
<span>
This means that:

</span>n = 60,011n₄ - 10,000 + 1,000*3 + 100*6

n = 60,011n₄ - 10,000 + 3,000 + 600

n = 60,011n₄ - 6,400

Therefore:

0<n₄<2, so n₄=1.

If n₄=1:

n = 60,011 - 6,400

n = 53,611

Answer:

53,611
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3 years ago
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Solve the system of linear equations by substitution.<br> y=X-4<br> - 2x+y= 18
11Alexandr11 [23.1K]

Answer:

] y=x-4

-2x+y=18

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y - 2x = 18

equation [2] for the variable y

[2] y = 2x + 18

// Plug this in for variable y in equation [1]

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[1] x = -22

// Solve equation [1] for the variable x

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// By now we know this much :

y = 2x+18

x = -22

// Use the x value to solve for y

y = 2(-22)+18 = -26

Step-by-step explanation:

please mark me as brainlist please

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