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3 years ago

If 80% of M is equal to 50% of N and N≠0, what is N M equal to?

Mathematics

1 answer:

asambeis [7]3 years ago

3 0

Answer:

0.625N

Step-by-step explanation:

0.50N

0.80M = 0.50N, so M = ------------

0.80

and so:

0.50N

N*M = N*------------ = (5/8)N, or 0.625N

0.80

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An above swimming pool is leaking after 1/2 hour the pool has leaked 7/8 of a gallon of water

Luden [163]

The answer to your question are <span>1.75</span>

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3 years ago

Which inequalities have the solution set graphed on the number line? Select two options.

Nat2105 [25]

Answer:

A,C,D

Step-by-step explanation:

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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 $y>2$

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7 0

3 years ago

10 points

marishachu [46]

Answer:

150+12x=630

Step-by-step explanation:

The x represents the amount of months and it is multiplied by 12 because each month costs 12$. 150 have to be paid up front and this added to the monthly fee is equal to 630$

3 0

3 years ago

What is the sequence of transformations that maps △ABC to △A′B′C′ ? Select from the drop-down menus to correctly identify each s

torisob [31]

<h3>Answer:</h3>

Any 1 of the following transformations will work. There are others that are also possible.

translation up 4 units, followed by rotation CCW by 90°.
rotation CCW by 90°, followed by translation left 4 units.
rotation CCW 90° about the center (-2, -2).

<h3>Step-by-step explanation:</h3>

The order of vertices ABC is clockwise, as is the order of vertices A'B'C'. Thus, if reflection is involved, there are two (or some other even number of) reflections.

The orientation of line CA is to the east. The orientation of line C'A' is to the north, so the figure has been rotated 90° CCW. In general, such rotation can be accomplished by a single transformation about a suitably chosen center. Here, we're told there is <em>a sequence of transformations</em> involved, so a single rotation is probably not of interest.

If we rotate the figure 90° CCW, we find it ends up 4 units east of the final position. So, one possible transformation is 90° CCW + translation left 4 units.

If we rotate the final figure 90° CW, we find it ends up 4 units north of the starting position. So, another possible transformation is translation up 4 units + rotation 90° CCW.

Of course, rotation 90° CCW in either case is the same as rotation 270° CW.

_____

We have described transformations that will work. What we don't know is what is in your drop-down menu lists. There are many other transformations that will also work, so guessing the one you have available is difficult.

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3 years ago