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Archy [21]
3 years ago
14

Jayden invested $22,000 in an account paying an interest rate of 2.5% compounded

Mathematics
1 answer:
Gemiola [76]3 years ago
3 0

Answer:

12.9

Step-by-step explanation:

30400=

30400=

\,\,22000e^{0.025t}

22000e

0.025t

Plug in

\frac{30400}{22000}=

22000

30400

​

=

\,\,\frac{22000e^{0.025t}}{22000}

22000

22000e

0.025t

​

Divide by 22000

1.3818182=

1.3818182=

\,\,e^{0.025t}

e

0.025t

\ln\left(1.3818182\right)=

ln(1.3818182)=

\,\,\ln\left(e^{0.025t}\right)

ln(e

0.025t

)

Take the natural log of both sides

\ln\left(1.3818182\right)=

ln(1.3818182)=

\,\,0.025t

0.025t

ln cancels the e

\frac{\ln\left(1.3818182\right)}{0.025}=

0.025

ln(1.3818182)

​

=

\,\,\frac{0.025t}{0.025}

0.025

0.025t

​

Divide by 0.025

12.9360062=

12.9360062=t

t = 12.9

12.9

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What is the middle term of the product of (x + 5)(x + 2)?
NeTakaya
When you multiply by distribution, the answer is: x^2 + 7x + 10. the middle term is 7x
5 0
4 years ago
Is this parallel, perpendicular or neither <br> y=4x + 3 and 4y + x = -1​
serious [3.7K]

<u>perpendicular lines</u>

Find the equation of the line that is

perpendicular to y = −4x + 10

and passes though the point (7,2)

The slope of y=−4x+10 is: −4

The negative reciprocal of that slope is:

m = −1−4 = 14

So the perpendicular line will have a slope of 1/4:

y − y1 = (1/4)(x − x1)

And now put in the point (7,2):

y − 2 = (1/4)(x − 7)

And that answer is OK, but let's also put it in "y=mx+b" form:

y − 2 = x/4 − 7/4

y = x/4 + 1/4

Answer:

I think it's neither parallel nor perpendicular.

step by step explanation:

  • Two lines will be parallel when their slopes are equal, and two lines will be perpendicular when their slopes are negative reciprocals of each other. Our slopes for these two equations are the coefficient for the x value. <u>Both slopes are </u>not equal so these lines are not parallel.
  • <u>Perpendicular Lines</u>

Two lines are Perpendicular when they meet at a right angle (90°).

3 0
3 years ago
6.3c-2(1.5c+4.1) simplify
svp [43]
6.3c-2(1.5c+4.1)=6.3c-3c-8.2=3.3c-8.2 :))))

7 0
3 years ago
Ben was walking in Manhattan at a constant speed. He went along 5th avenue from 4th to 98th street. At 3:00 he was on 15th stree
muminat

Answer: The starting time is 2:27 and finishing time is 7:09.    


Step-by-step explanation: It is given that Ben started from 4th street and he finished at 98th street.

Also, at 3:00, he was on 15th street and at 4:30, he was on 45th street.

That is, time taken to cover (45-15), i.e., 30 streets is 90 minutes,                           so the time taken to cover 1 street is 3 minutes.

Therefore, Ben covers distance from one street to second in 3 minutes. Since he started from 4th street, and there are 11 streets to cover between 4 and 15, so Ben's starting time was (3:00 - 3×11 min) = 2:27.

And his finishing time was (4:30 + 3×53 min) = 7:09.

Again, the equation that tells us on what street 'N' he was after time 'T' of his starting can be written as

N=4+\dfrac{T}{3.}

Thus, the starting and finishing time was 2:27 and 7:09 respectively.



3 0
3 years ago
Write the equation of a parabola whose directrix is y=9.75 and has a focus at (8, 0.25).
lana [24]

Answer:

y = (-1/19)(x - 8)² + 1521/76

Step-by-step explanation:

A parabola moves in such a way that it's distance from it's focus and directrix are always equal.

Now, we are given that directrix is y = 9.75 and focus is at (8, 0.25). Focus can be rewritten as (8, ¼) and directrix can be rewritten as y = 39/4

If we consider a point with the coordinates (x, y), it means the distance from this point to the focus is;

√((x - 8)² + (y - ¼)²)

Distance from that point to the directrix is; (y - 39/4)

Thus;

√((x - 8)² + (y - ¼)²) = (y - 39/4)

Taking the square of both sides gives;

((x - 8)² + (y - ¼)²) = (y - 39/4)²

(x - 8)² + y² - ½y + 1/16 = y² - (39/2)y + (39/4)²

Simplifying this gives;

(x - 8)² - (39/4)² = (½ - 39/2)y

(x - 8)² - 1521/4 = -19y

(x - 8)² - 1521/4 = -19y

Divide both sides by -19 to get;

y = (-1/19)(x - 8)² + 1521/76

6 0
3 years ago
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