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3 years ago

Please show all of the work ok

Mathematics

1 answer:

katen-ka-za [31]3 years ago

3 0

3. 2.5. Take 25/10 and you get 2.5

4. $406.00. (580*30) = 17400/100 = 174, 580-174 = $406

5. $425.00. (500*15) = 7500/100 = 75, 500-75 = $425

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A deposit of $7,000 at 6.5% for 120 days. 151.67 358.97

astraxan [27]

120 days is approximately equals to 4 months. (30 days * 4 = 120 days)
Now you deposited 7000 dollars at 6.5% interests.
IN 1 months
=> 7000 * .065 = 455 dollars in 1 month is the interest
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4 years ago

Assume log bx = 0.49. Evaluate. Logbx^2

Snowcat [4.5K]

Step-by-step explanation:

Use exponent property of logs.

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3 years ago

Write a factored equation that has a solution set of {-12, 15}

Alexus [3.1K]

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(x+12)(x-15)

Step-by-step explanation:

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3 years ago

Can anyone help me with this problem

dimaraw [331]

the answer would be a= -1

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1 year ago

HALLP QUICKKKK

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 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:
 $S_{a}=S_{l}-2$
We also know that in the lake the boat sailed for 1 hour longer than it sailed in the river, so:
 $t_{l}=t_{a}+1$

Now, we can set up our equations.
Speed of the boat traveling in the river:
 $S_{a}= \frac{6}{t_{a} }$
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.

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4 years ago