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Nadusha1986 [10]
3 years ago
12

1. Which equation does not represent a

Mathematics
2 answers:
aivan3 [116]3 years ago
6 0
B) y=x

isn’t a linear equation because it has to be written like y=mx+b
Murrr4er [49]3 years ago
6 0
Hdjsjskskdlslsndndbdkskndd
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Enter an algebraic expression to model the given context. Give your answer in simplest form. The original price p of an item les
ra1l [238]

Answer:

Step-by-step explanation:

Let

Original price = p

Less Discount = 20%

Total discount paid for the item = 20% of p

= 20/100 * p

= 0.2 * p

= 0.2p

Actual price after discount = 100% - 20%

= 80%

Total price paid for the item = 80% of p

= 80/100 * p

= 0.8 * p

= 0.8p

6 0
3 years ago
State whether the given measurements determine zero, one, or two triangles.
Lady bird [3.3K]
I bet you're doing the Law of Sines right now. If you use the law of sines you find out that angle A is a 90 degree angle. There is only one triangle you can make with these measurements because if you have one angle of 30 and another of 90, the 3rd angle has to be a 60 degree angle. So it is a special right triangle, 30-60-90
8 0
3 years ago
Heyyyyy! Need Help with this! I will give brainliest to correct answer!! <3
Alex787 [66]

Answer:

The answer is 1/8 pints left

Step-by-step explanation:

If you had 3/4 pints already and you poured 5/8 pints into your glass, you would only have 1/8 pints left. First you need to make them both have the same denominator (8) to do that you need to multiply 3/4(2)=6/8. 6/8-5/8=1/8 pints left.

5 0
2 years ago
Find the length of side a.
coldgirl [10]
The length of side A is 194
6 0
2 years ago
Read 2 more answers
A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,
guajiro [1.7K]

We have been given that a huge ice glacier in the Himalayas initially covered an area of 45 square kilo-meters. The relationship between A, the area of the glacier in square kilo-meters, and t, the number of years the glacier has been melting, is modeled by the equation A=45e^{-0.05t}.

To find the time it will take for the area of the glacier to decrease to 15 square kilo-meters, we will equate A=15 and solve for t as:

15=45e^{-0.05t}

\frac{15}{45}=\frac{45e^{-0.05t}}{45}

\frac{1}{3}=e^{-0.05t}

Now we will switch sides:

e^{-0.05t}=\frac{1}{3}

Let us take natural log on both sides of equation.

\text{ln}(e^{-0.05t})=\text{ln}(\frac{1}{3})

Using natural log property \text{ln}(a^b)=b\cdot \text{ln}(a), we will get:

-0.05t\cdot \text{ln}(e)=\text{ln}(\frac{1}{3})

-0.05t\cdot (1)=\text{ln}(\frac{1}{3})

-0.05t=\text{ln}(\frac{1}{3})

t=\frac{\text{ln}(\frac{1}{3})}{-0.05}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-0.05\cdot 100}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-5}

t=-\text{ln}(\frac{1}{3})\cdot 20

t=-(\text{ln}(1)-\text{ln}(3))\cdot 20

t=-(0-\text{ln}(3))\cdot 20

t=20\text{ln}(3)

Therefore, it will take 20\text{ln}(3) years for area of the glacier to decrease to 15 square kilo-meters.

6 0
3 years ago
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