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notka56 [123]
3 years ago
14

8 m 3 m 5 m 2 m 3 m 9 m 8 m

Mathematics
1 answer:
Andrew [12]3 years ago
5 0

Answer:

this makes no sense at all, what's the question?

You might be interested in
If (x − 2) ∶ 5 = 7 ∶ 9, then find the value of x.
WARRIOR [948]

Answer:

x=53/9

Step-by-step explanation:

(x-2):5=7:9

9x-18=35

9x=35+18

9x=53

x=53/9

4 0
3 years ago
-12 + 2a = -6(1 + a) + 8a
valkas [14]
-12 + 2a = -6 (1+a)?
Answer is 0 = 6
8 0
2 years ago
How do I find the slope of the line through (–9, –10) and (–2, –5)?
storchak [24]
Answer:
slope of line is 5/7

Explanation:
The slope of the line can be calculated using the following rule:
slope = \frac{y2-y1}{x2-x1}

The given points are:
(-9,-10) which represent (x1,y1)
(-2,-5) which represent (x2,y2)

Substitute with the points in the above equation to get the slope as follows:
slope = \frac{-5-(-10)}{-2-(-9)} = 5/7

Hope this helps :)

4 0
4 years ago
Read 2 more answers
What is the probability of drawing a 9 of heart or an ace from a standard deck of cards? A. 7/52 B. 5/52 C. 5/26 D. 8/13 What is
Strike441 [17]
A for the first because  there are 4 aces and 1 nine of hearts. that makes it 5 cards of the 52 in a deck
d for the second because there are two cards: the queen of heart and and ace of diamond out of the 52

hope this helped :)

5 0
3 years ago
What is the range of the equation
a_sh-v [17]

The range of the equation is y>2

Explanation:

The given equation is y=2(4)^{x+3}+2

We need to determine the range of the equation.

<u>Range:</u>

The range of the function is the set of all dependent y - values for which the function is well defined.

Let us simplify the equation.

Thus, we have;

y=2 \cdot 4^{x+3}+2

This can be written as y=2^{1+2(x+3)}+2

Now, we shall determine the range.

Let us interchange the variables x and y.

Thus, we have;

x=2^{1+2(y+3)}+2

Solving for y, we get;

x-2=2^{1+2(y+3)}

Applying the log rule, if f(x) = g(x) then \ln (f(x))=\ln (g(x)), then, we get;

\ln \left(2^{1+2(y+3)}\right)=\ln (x-2)

Simplifying, we get;

(1+2(y+3)) \ln (2)=\ln (x-2)

Dividing both sides by \ln (2), we have;

2 y+7=\frac{\ln (x-2)}{\ln (2)}

Subtracting 7 from both sides of the equation, we have;

2 y=\frac{\ln (x-2)}{\ln (2)}-7

Dividing both sides by 2, we get;

y=\frac{\ln (x-2)-7 \ln (2)}{2 \ln (2)}

Let us find the positive values for logs.

Thus, we have,;

x-2>0

     x>2

The function domain is x>2

By combining the intervals, the range becomes y>2

Hence, the range of the equation is y>2

7 0
3 years ago
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