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trasher [3.6K]
3 years ago
15

If k(x) = 5x - 6, which expression is equivalent to (k + k)(4)?

Mathematics
2 answers:
ycow [4]3 years ago
5 0

\large \boxed{(k + k)(4) = k(4) + k(4)}

First, we find the value of k(4). We can find the value of k(4) by substituting x = 4 in k(x).

\large{k(x) = 5x - 6  \longrightarrow k(4) = 5(4) - 6} \\  \large{k(4) = 20 - 6 \longrightarrow 14}

Therefore the value of k(4) is 14. Next we find k(4)+k(4) which is equivalent to 14+14. Hence, the value of (k+k)(4) is 28. There's also another method.

\large \boxed{(k + k)(4) = 2k(4)}

As we know k(4) is 14. Therefore 2k(4) is 2×14 = 28.

Answer

  • (k+k)(4) = 28

kogti [31]3 years ago
3 0
28! hope this helps, you just plug in 4 for x, and then multiply by two.
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Find x show ur work !!
Elis [28]

Answer:

x = 10

Step-by-step explanation:

A triangle's angles add up to 180 degrees, so 180-90 (the right angle) is 90 degrees. That means that

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4 0
3 years ago
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If you were to roll the dice one time what is the probability it will NOT land on a 2? Put your answer in fraction form.
OLEGan [10]

Answer:

5/6

Step-by-step explanation:

On a standard 6-sided die, there is only one 2. That means that there are 5 other numbers that it could land on (1, 3, 4, 5, 6).

Using that information, the probability of it not landing on a 2 is 5 out of 6 or 5/6. This is because you must do part over whole. The "part" in this situation is the 5 "wanted" numbers, and the "whole" is 6 because there are six potential numbers that it could land on.

I hope this helps.

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Answer:

The probability that it will still be working after one week is \frac{1}{5}

Step-by-step explanation:

Given :

Total number of bulbs = 25

Number of bulbs which are good condition and will function for at least 30 days = 5

Number of bulbs which are partially defective and will fail in their second day of use = 10

Number of bulbs which are totally defective and will not light up = 10

To find : What is the probability that it will still be working after one week?

Solution :

First condition is a randomly chosen bulb initially lights,

i.e. Either it is in good condition and partially defective.

Second condition is it will still be working after one week,

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So, favorable outcome is 5

The probability that it will still be working after one week is given by,

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

\text{Probability}=\frac{5}{25}

\text{Probability}=\frac{1}{5}

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Answer:

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