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Juli2301 [7.4K]
3 years ago
14

What is the area of the triangle below?

Mathematics
1 answer:
Cloud [144]3 years ago
7 0
The area of the triangle below is 7.5 units
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Ella has a box of chocolate candies. She gives 1/3 of the candies to her sister, 4 to her brother, and eats the remaining 12 can
tatyana61 [14]

Answer:

24 candies originally

Step-by-step explanation:

Add the 4 to how many you had left, then think about that number, 16, and what goes into 16 evenly 2 times? 8 and originally she gave away 1/3 of the box and if you were to add another 8 to the box it would be 24 divided by 1/3, -,-4,= 12.

3 0
3 years ago
Please help me
Snowcat [4.5K]

Answer:

Option 1, 2, and 4 are correct. Option 3 is incorrect.

6 0
2 years ago
Lucy rented a movie. She started the movie at 9:48 a.m. and it was 3 hours 39 minutes long. When did the movie end
iren2701 [21]

Think through this one...

9:48 + 9 minutes = 9:57

9:57 + 30 minutes = 10:27

10:27 + 3 hours = 1:27 p.m.

6 0
3 years ago
Read 2 more answers
Solving quadratic equation<br>(2m+3)(4m+3)=0
Basile [38]
8m^2+18m+9=0

Multiply 2m by 4m to get 8m^2. Multiply 2m by 3 to get 6m. Multiply 3 by 4m to get 12m. Add the 6m and 12m to make 18m. Multiply 3 by 3 to get the integer 9. Set it all equal to 0.
7 0
3 years ago
A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min
sesenic [268]
A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})
\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}
A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4

We're given that A(0)=30. Multiply both sides by the integrating factor e^{t/100}, then

e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4
\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4
e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C
A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}

Given that A(0)=30, we have

30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02

so the amount of salt in the tank at time t is

A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}
3 0
3 years ago
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