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3 years ago

8

Ponce Leon has a principal of $900 in his savings account on October 1. The money earns an APR of 6.5% calculated quarterly as s

imple interest. What is the amount in the account on July 1 of the following year?

1 answer:

alisha [4.7K]3 years ago

5 0

A=900×((1+(0.065÷4)×(9÷12)))
A=910.96

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Find tan0, cos0, and csc0, where 0 is the angle shown in the figure. Give exact values, not decimal approximations. BRANLIEST!!

Setler [38]

Answer:

see explanation

Step-by-step explanation:

We require to find the third side of the triangle.

Using Pythagoras' identity

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides.

let x represent the third side, then

x² + 15² = 17², that is

x² + 225 = 289 ( subtract 225 from both sides )

x² = 64 ( take the square root of both sides )

x = $\sqrt{64}$ = 8

Thus

tanΘ = $\frac{opposite}{adjacent}$ = $\frac{15}{8}$

cosΘ = $\frac{adjacent}{hypotenuse}$ = $\frac{8}{17}$

sinΘ = $\frac{opposite}{hypotenuse}$ = $\frac{15}{17}$

cscΘ = $\frac{1}{sin0}$ = $\frac{1}{\frac{15}{17} }$ = $\frac{17}{15}$

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3 years ago

Helppppp plsssss and make it fast

irga5000 [103]

Answer:

1 & 2 perpudicular

1 & 3 neither

2 & 3 is parallel

Step-by-step explanation:

7 0

3 years ago

Rina8888 [55]

To solve this we are going to use the formula for speed: $S= \frac{d}{t}$
where
$S$ is the speed
$d$ is the distance
$t$ is the time

Let $S_{l}$ be the speed of the boat in the lake, $S_{a}$ the speed of the boat in the river, $t_{l}$ the time of the boat in the lake, and $t_{a}$ the time of the boat in the river.

We know for our problem that <span>the current of the river is 2 km/hour, so the speed of the boat in the river will be the speed of the boat in the lake minus 2km/hour:
</span> $S_{a}=S_{l}-2$
We also know that in the lake the boat<span> sailed for 1 hour longer than it sailed in the river, so:
</span> $t_{l}=t_{a}+1$
<span>
Now, we can set up our equations.
Speed of the boat traveling in the river:
</span> $S_{a}= \frac{6}{t_{a} }$
But we know that $S_{a}=S_{l}-2$ , so:
$S_{l}-2= \frac{6}{t_{a} }$ equation (1)

Speed of the boat traveling in the lake:
$S_{l}= \frac{15}{t_{l} }$
But we know that $t_{l}=t_{a}+1$ , so:
$S_{l}= \frac{15}{t_{a}+1}$ equation (2)

Solving for $t_{a}$ in equation (1):
$S_{l}-2= \frac{6}{t_{a} }$
$t_{a}= \frac{6}{S_{l}-2}$ equation (3)

Solving for $t_{a}$ in equation (2):
$S_{l}= \frac{15}{t_{a}+1}$
$t_{a}+1= \frac{15}{S_{l}}$
$t_{a}=\frac{15}{S_{l}}-1$
$t_{a}= \frac{15-S_{l}}{S_{l}}$ equation (4)

Replacing equation (4) in equation (3):
$t_{a}= \frac{6}{S_{l}-2}$
$\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}$

Solving for $S_{l}$ :
$\frac{15-S_{l}}{S_{l}}=\frac{6}{S_{l}-2}$
$(15-S_{l})(S_{l}-2)=6S_{l}$
$15S_{l}-30-S_{l}^2+2S_{l}=6S_{l}$
$S_{l}^2-11S_{l}+30=0$
$(S_{l}-6)(S_{l}-5)=0$
$S_{l}=6$ or $S_{l}=5$

We can conclude that the speed of the boat traveling in the lake was either 6 km/hour or 5 km/hour.

3 0

4 years ago

Solve the following system of equations. 2x + y = 3 x = 2y - 1 Make sure there are NO SPACES in your answer. Include a comma in

mrs_skeptik [129]

Answer: $x=1;y=1$ or (1,1)

Step-by-step explanation:

You can apply the Method of substitution.

Knowing that $x = 2y - 1$ (second equation), you can substitute it into the first equation and solve for the variable y:

$2x+y=3\\2(2y - 1)+y=3\\4y-2+y=3\\5y=3+2\\5y=5\\y=1$

Now you need to substitute $y=1$ into the second equation $x = 2y - 1$ to calculate the value of the variable x. Then:

$x = 2(1) - 1\\x=2-1\\x=1$

Then, the solution is:

$x=1;y=1$ or (1,1)

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3 years ago

Read 2 more answers

Please help need an answer really fast

dimulka [17.4K]

It would be 1 and 6/7

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4 years ago