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Alborosie
3 years ago
15

In one year, Michael earned $6300 as a work study in college. He invested part of the money at 9% and the rest at 7%. If he rece

ived a total of $493 in interest at the end of the year, how much was invested at 7%? How much was invested 9%?
Mathematics
1 answer:
Nitella [24]3 years ago
3 0
Let x represent invested money at 9% and y at 7%
Now we write system:
x+y = 6300
1.09*x + 1.07* y = 6300 + 493 = 6793

1.09x + 1.09y = 6867
Here we multiplied first equation with 1.09 and now we will subtract second from first:

0.02y = 74
y = 3700
x = 6300-3700 = 2600

Michael invested 2600 at 9% and 3700 at 7%
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What is the horizontal asympote of this graph?
Anna35 [415]

Using it's concept, it is found that the graph has no horizontal asymptote.

<h3>What are the horizontal asymptotes of a function f(x)?</h3>

The horizontal asymptote is the value of f(x) as x goes to infinity, as long as this value is different of infinity.

In this problem, we have that:

  • The function is undefined for x < 0, hence \lim_{x \rightarrow -\infty} f(x) is undefined.
  • For x > 0, the funciton goes to infinity, hence \lim_{x \rightarrow \infty} f(x) = \infty.

Thus, the graph has no horizontal asymptote.

More can be learned about horizontal asymptotes at brainly.com/question/16948935

#SPJ1

4 0
2 years ago
The measure of _A is 18° greater than the measure of _B. The two angles are complementary. Find the
kozerog [31]

Answer:

angle A=54 degree

angle B =36 degree

Step-by-step explanation:

let angle B be x

angle A=x+18

since they are complementary angles sum of these two angles will be 90 degree

x+x+18=90

2x=90-18

2x=72

x=72/2

x=36 degree

substitute the value of x to find angle A and angle B

for angle A

x+18

36+18

54 degree

for angle B

angle B =x

=36 degree

4 0
3 years ago
$2,000 IN AN ACCOUNT THAT PAID 6.25% SIMPLE INTEREST ANNUALLY
valkas [14]
\bf ~~~~~~ \textit{Simple Interest Earned Amount}\\\\&#10;A=P(1+rt)\qquad &#10;\begin{cases}&#10;A=\textit{accumulated amount}\\&#10;P=\textit{original amount deposited}\to& \$2000\\&#10;r=rate\to 6.25\%\to \frac{6.25}{100}\to &0.0625\\&#10;t=years\to &3&#10;\end{cases}&#10;\\\\\\&#10;A=2000(1+0.0625\cdot 3)\implies A=2000(1.1875)
8 0
3 years ago
What is .08333333333 as a fraction ?
madam [21]
<span>.0833333333 with repeating 3 is (2/24) or (1/12) <- There's you're answer.

<em>(Just put this up here since I answered it in a comment)</em></span>
5 0
3 years ago
Find the limit, if it exists. (If an answer does not exist, enter DNE.) lim (x, y) → (0, 0) x4 − 34y2 x2 + 17y2
HACTEHA [7]

Answer:

<h2>DNE</h2>

Step-by-step explanation:

Given the limit of the function \lim_{(x,y) \to (0,0)} \frac{x^4-34y^2}{x^2+17y^2}, to find the limit, the following steps must be taken.

Step 1: Substitute the limit at x = 0 and y = 0 into the function

= \lim_{(x,y) \to (0,0)} \frac{x^4-34y^2}{x^2+17y^2}\\=  \frac{0^4-34(0)^2}{0^2+17(0)^2}\\= \frac{0}{0} (indeterminate)

Step 2: Substitute y = mx int o the function and simplify

= \lim_{(x,mx) \to (0,0)} \frac{x^4-34(mx)^2}{x^2+17(mx)^2}\\\\= \lim_{(x,mx) \to (0,0)} \frac{x^4-34m^2x^2}{x^2+17m^2x^2}\\\\\\= \lim_{(x,mx) \to (0,0)} \frac{x^2(x^2-34m^2)}{x^2(1+17m^2)}\\\\\\= \lim_{(x,mx) \to (0,0)} \frac{x^2-34m^2}{1+17m^2}\\

= \frac{0^2-34m^2}{1+17m^2}\\\\=  \frac{34m^2}{1+17m^2}\\\\

<em>Since there are still variable 'm' in the resulting function, this shows that the limit of the function does not exist, Hence, the function DNE</em>

4 0
3 years ago
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