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Akimi4 [234]
3 years ago
7

HELP PLEASE HELP AND HURRY

Mathematics
1 answer:
VikaD [51]3 years ago
6 0

Answer:

Alguien me ayudaaa por faa

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3 years ago
PLS HELP WILL GIVE BRAINLIEST
andreyandreev [35.5K]

The guts of the table are ...

\left[\begin{array}{cccc}x&y&x+y=85&y=2x+4\\19&65&\text{false}&\text{false}\\25&60&\text{true}&\text{false}\\27&58&\text{true}&\text{true}\\32&53&\text{true}&\text{false}\end{array}\right]

Of course, you know the answer after figuring the third line of the table.

The two numbers are 27 and 58.

_____

If the second number is odd, the result after subtracting 4 will not be divisible by 2. This lets you reject the 1st and 4th choices immediately. As for the second choice, twice 25 is 10 less than 60, not 4 less, so that choice can also be discarded.

Starting from the system of equations

x+y=85\\y=2x+4

you can use substitution to get

x+(2x+4)=85\\\\3x=81\qquad\text{subtract 4}\\\\x=27\qquad\text{divide by 3}

This identifies the 3rd selection as the appropriate one.

5 0
3 years ago
Read 2 more answers
How to do this?? give me a solution its confusing me?!!​
Alika [10]

Answer:

t = 3h

Step-by-step explanation:

Given

h = \frac{1}{3} t

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3h = t

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4 years ago
Expand and simplyfy (4x+3)(2x-5)
zheka24 [161]
(4x+3) (2x-5)
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3 years ago
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The expression is: | x | = A + B + (A < C < B)

To simplify the expression, write without the absolute value sign and we would get this:

<span>x = A + B + (A < C < B)</span>

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