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lubasha [3.4K]
2 years ago
6

What is the point of a pythagorean theorem?

Mathematics
1 answer:
Nikitich [7]2 years ago
3 0

The Pythagorean Theorem describes an important geometric relationship between the three sides of a right triangle.

First of all, to understand the Pythagorean Theorem, you need to know about angles and triangles, and you also need to know about exponents and square roots. These things will help you understand how it works and how to solve for a missing side.

The Pythagorean Theorem

For a right triangle

with legs a and b and the hypotenuse c

a^{2} + b^{2} = c^{2}

The Pythagorean Theorem does not apply to all triangles.

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PLEASE ANSWER FAST!<br> you dont have to explain but im trusting on u have a good day
JulsSmile [24]

Answer:

126 square inches

Step-by-step explanation:

The area of the figure = area of the parallelogram + area of the square + area of the rectangle

✔️Area of the parallelogram = b*h

b = 10 in.

h = 3.5 in.

Area = 10*3.5

Area = 35 in.²

✔️Area of the square = s²

s = 4 in.

Area = 4²

Area = 16 in.²

✔️Area of the rectangle = L*W

L = 12.5 in.

W = 6 in.

Area = 12.5*6

Area = 75 in.²

✅Area of the figure = 35 + 16 + 75

= 126 in.²

8 0
2 years ago
Which statement is true about ∠1 and ∠2?
Serjik [45]
The answer is C.

All of the other answers are incorrect
5 0
3 years ago
Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite
zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt  be a gram/L

Let the amount salt in the tank at any time t be Q(t).

\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}

Incoming rate = (a g/L)×(1 L/min)

                       =a g/min

The concentration of salt in the tank at any time t is = \frac{Q(t)}{10}  g/L

Outgoing rate = (\frac{Q(t)}{10} g/L)(1 L/ min) \frac{Q(t)}{10} g/min

\frac{dQ}{dt} = a- \frac{Q(t)}{10}

\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt

Integrating both sides

\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt

\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c        [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c

\Rightarrow -log|10a-15|-2=c

Therefore ,

-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2 .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0  in equation (1) we get

- log|10a|= -log|10a-15| -2

\Rightarrow- log|10a|+log|10a-15|= -2

\Rightarrow log|\frac{10a-15}{10a}|= -2

\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}

\Rightarrow 1-\frac{15}{10a} =e^{-2}

\Rightarrow \frac{15}{10a} =1-e^{-2}

\Rightarrow \frac{3}{2a} =1-e^{-2}

\Rightarrow2a= \frac{3}{1-e^{-2}}

\Rightarrow a = 1.73

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0
3 years ago
To make concrete it takes 8 parts crushed gravel, 7 parts water and 5 parts cement mix. A worker needs to make 220kg of concrete
Alika [10]
A) 8 + 7 + 5 = 20 parts
b) 220 kg ÷ 20 = 11
c) <span> 8 parts crushed gravel: 8 x 11 = 88 kg
     7 parts water: 7 x 11 = 77 kg
     5 parts cement: 5 x 11 = 55 kg</span>
8 0
2 years ago
Lucy made $65 for 5 hours of work.
shusha [124]

Answer:

169

Step-by-step explanation:

65 divided by 5 to get 13

and multiply that with 13

like this: 13x13

to get 169

6 0
2 years ago
Read 2 more answers
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