Enter the sum of numbers as a product of their GCF.75 + 90

If I 50% increase is followed by a 33 1\3 % decrease is it greater, less than, or the same as the original

luda_lava [24]

Answer:

Greater than.

Step-by-step explanation:

Let the original value be x.

A 50% increase in x is given by

x + 50%x

=> x + 0.5x

=> 1.5x

Followed by a 33 $\frac{1}{3}$ % decrease in the value gives:

1.5x - 33 $\frac{1}{3}$ %(1.5x)

=> 1.5x - 0.33(1.5x)

=> 1.5x - 0.495x

=> 1.005x

Since 1.005x is greater than the original value, x, then:

A 50% increase followed by a 33 $\frac{1}{3}$ % decrease gives a value that is greater than the original.

5 0

3 years ago

Ramer and knox began a partnership by investing $64,000 and $94,000, respectively. the partners agreed to share net income and l

Butoxors [25]

Each partner's share given a first-year net income of $102,800 is $-3,300.

Each partner's share given a first-year net loss of $20,800 is $-44,300.

<h3>What is each partner's share?</h3>

Partner' share = net income / loss - interest - salary salary allowance

Interest = ($64000 x 10%) + ($94000x 10%) = $15,800

Salary = $52,000 + $41,600 = $93,600

Share of net income = [$102,800 - (15,800 + 93600)] / 2 = $-3,300

Share of net loss = [-20,800 - (15,800 + 93600)] / 2 = $-44,300

To learn more about profit, please check: brainly.com/question/26181966

5 0

2 years ago

Can you help me with these 4 problems?

adoni [48]

What are the questions attach a pic plz

3 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

A function f (x)=-8x^2. What is f (-3)?

Sveta_85 [38]

F(x)= -8x²

f(-3)= -8* (-3)²

f(-3)= -8 * 9

f(-3)= -72

8 0

3 years ago

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