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2 years ago

How does Elian's drawing show that the number 10 is a composite number?

Mathematics

1 answer:

kumpel [21]2 years ago

6 0

It shows that there is only one way to get to 10 which is by multiplying it by 1

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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 $y>2$

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2 years ago

8x-29 + 6x-11 + 3x-1 solve for x

elena-s [515]

Answer:

Step-by-step explanation:

I need help on this too.

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5c + 9c <br><br>Show your solution! tyms

S_A_V [24]

5c plus 9c is 14c bc 5 plus 9 is 14 and the exponent is c so 14c

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Bob has some 10 ltb weights and some 3 lb weights. Together, all his weights add up to 50 lbs. If x represents the number of 3 l

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total weight = 50

3x + 10y = 50

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2 years ago

You have decided to purchase a new computer with the Amtel processor. The new processor is the latest release of

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A. 3,535.20

All you gotta do is Multiply 36 with 98.20 and boom, you got the answer

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