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3 years ago

What is the balance of 6,000 compounded annually at a rate of 4% for 10 years

Mathematics

2 answers:

VashaNatasha [74]3 years ago

7 0

Answer:

the final balance is $8,945

Vitek1552 [10]3 years ago

5 0

Answer:

8,881.41

Step-by-step explanation:

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Write an equation in the form y = mx + b for the line which contains the points (8, 25) and (1, 4).

MaRussiya [10]

$\frac{25-4}{8-1}= \frac{21}{7} =3$ . So the slope is 3 which you can plug in one of the points and get the final equation which is : $y=3x+1$

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3 years ago

How many solutions does the equation have??<br> 3(d+11)=6(d+33) PLEASE HELP MEEEEEEE

Misha Larkins [42]

A powerful way to find out is to solve the equation. Let's try that.

   3 (d + 11) = 6 (d + 33)

Eliminate the parentheses: 3d + 33 = 6d + 198

Subtract 3d from each side: 33 = 3d + 198

Subtract 198 from each side: -165 = 3d

Divide each side by 3 : - 55 = d

There it is ... the solution to the original equation.
'd' has to be -55, otherwise the original equation isn't a true statement.
-55 is the ONLY number that 'd' can be. If you write in any other number
for 'd', the equation will be false.

For example, less see what the equation says when d=2 :

   3(2 + 11) = 6(2 + 33)

   3( 13 ) = 6( 35 )

   39    = 210

Is that a true statement ? Is 39 equal to 210 ?
No. 39 and 210 are not equal. They are different.
So the equation is not true when d = 2 .
The equation is true ONLY when d = -55 .

The equation has exactly one solution.

4 0

3 years ago

What is 127/100 as a percent

jenyasd209 [6]

1.27, think money.
100 pennies is $1
27 cents left
$1.27

3 0

4 years ago

Read 2 more answers

Solve the quadratic equation.<br>m^2/15 -3 = 2

Radda [10]

The answer is $5 \sqrt{3}$

<h3>Explanation :</h3>

$\frac{ {m}^{2} }{15} - 3 = 2$

$\frac{ {m}^{2} }{15} = 2 + 3$

$\frac{ {m}^{2} }{15} = 5$

${m}^{2} = 5 \times 15$

${m}^{2} = 75$

$m = \sqrt{75}$

$m = \sqrt{25 \times 3}$

$m = \sqrt{25} \times \sqrt{3}$

$m = 5 \sqrt{3}$

CMIIW

7 0

3 years ago

Question 6

Firdavs [7]

So hmmmmm, notice... the boat went up in 3hrs, came back to the starting point in 2hrs, it went up 108km, it came back, well, from 108km to distance 0, so the distance on the way back is just the same 108km

now... let's say the stream has a speed rate of "r", and the boat has a still water speed rate of "b"

bear in mind that, when the boat is going UP, is not really going "b" fast, because the stream's "r" rate is going against it, and thus subtracting "r" from "b", so is really going " b - r " fast

when the boat is going down, is not going "b" fast either, because, again the stream's rate "r" is adding to it, because is going with the current, so is really going " b + r " fast

now, recall your d =rt, distance = rate * time

$\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
upstream&108&b-r&3\\
downstream&108&b+r&2
\end{array}\\\\
-------------------------------\\\\
\begin{cases}
108=3(b-r)\implies \frac{108}{3}=b-r\\
\qquad 36=b-r\implies 36+r=\boxed{b}\\\\
108=2(b+r)\implies 54=b+r\\
\qquad 54=\boxed{36+r}+r
\end{cases}$

solve for "r"

what's b? well, 36+ r = b

7 0

4 years ago