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Dennis_Churaev [7]
2 years ago
14

ASAP PLS!! -GIVING OUT BRAINLIEST

Mathematics
1 answer:
Korvikt [17]2 years ago
8 0

Answer:

The answer is (C)

Step-by-step explanation:

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I need this done by tonight pls help!
vovangra [49]

Answer:

8w-16

Step-by-step explanation:

8(w-2)

8(w) & 8(-2)

8w-16

7 0
3 years ago
I need help with this one math question please!
den301095 [7]

Answer:

see explanation

Step-by-step explanation:

Assuming the fractions are being multiplied

Factorise the denominators of both fractions

x² - 3x - 10 = (x - 5)(x + 2)

x² + x - 12 = (x + 4)(x - 3)

The product can now be expressed as

\frac{x-5}{(x-5)(x+2)} × \frac{x+2}{(x+4)(x-3)}

Cancel (x - 5) and (x + 2) on the numerators/ denominators, leaving

\frac{1}{(x+4)(x-3)} = \frac{1}{x^2+x-12}

3 0
3 years ago
Whats the graph after you transition it 6 units left?
viva [34]

S: (-6,2)

R: (3,2)

P: (-4,-6)

Q: (-2,-6)

Hope this helped!

-TTL

7 0
3 years ago
PLEASE HELP WILL MARK BRAINLIEST
Sholpan [36]

Answer: Choice D

b greater-than 3 and StartFraction 2 over 15 EndFraction

In other words,

b > 3 & 2/15

or

b > 3\frac{2}{15}\\\\

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

Explanation:

Let's convert the mixed number 2 & 3/5 into an improper fraction.

We'll use the rule

a & b/c = (a*c + b)/c

In this case, a = 2, b = 3, c = 5

So,

a & b/c = (a*c + b)/c

2 & 3/5 = (2*5 + 3)/5

2 & 3/5 = (10 + 3)/5

2 & 3/5 = 13/5

The inequality 2 \frac{3}{5} < b - \frac{8}{15}\\\\ is the same as \frac{13}{5} < b - \frac{8}{15}\\\\

---------------------

Let's multiply both sides by 15 to clear out the fractions

\frac{13}{5} < b - \frac{8}{15}\\\\15*\frac{13}{5} < 15*\left(b - \frac{8}{15}\right)\\\\39 < 15b-8\\\\

---------------------

Now isolate the variable b

39 < 15b-8\\\\15b-8 > 39\\\\15b > 39+8\\\\15b > 47\\\\b > \frac{47}{15}\\\\b > \frac{45+2}{15}\\\\b > \frac{45}{15}+\frac{2}{15}\\\\b > 3+\frac{2}{15}\\\\b > 3\frac{2}{15}\\\\

Side note: Another way to go from 47/15 to 3 & 2/15 is to notice how

47/15 = 3 remainder 2

The 3 is the whole part while 2 helps form the fractional part. The denominator stays at 15 the whole time.

7 0
3 years ago
What is the range of the function below in set builder notation? y=|x-3|
Kisachek [45]

Answer:  \{y | \ y \in \mathbb{R}, \ y \ge 0\}

This translates to "y is any real number such that it is 0 or larger".

The reasoning is that the result of any absolute value function is either 0 or positive. In other words, we'll never get a negative result of an absolute value function. This is due to how absolute value represents distance. Negative distance does not make sense.

So if y = |x-3| then y = 0 is the smallest output possible. We could have any positive output we want.

In terms of a graph (see below), the V shape is at the lowest point (3,0). The y coordinate is all we care about in terms of finding the range. So we see the lowest y value is y = 0.

4 0
3 years ago
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