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Snowcat [4.5K]
2 years ago
5

How do you do simultaneous equations by the method or substitution?​

Mathematics
1 answer:
ozzi2 years ago
5 0

Answer:

Here's the simplest example possible: let's say x + y = 3 and x - y = 1. Solve the second equation for x by adding y to both sides: (x - y) + y = 1 + y. So x = 1 + y. Take that value of x, and substitute it into the first equation given above (x + y = 3).

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Find the solutions to 6x^2- 54x = 0
Nina [5.8K]

Answer:      

A. x = 9

D. x = 0

Step-by-step explanation:

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have            

6x^{2} -54x=0              

so

a=6\\b=-54\\c=0

substitute in the formula

x=\frac{-(-54)(+/-)\sqrt{-54^{2}-4(6)(0)}} {2(6)}

x=\frac{54(+/-)54} {12}

x=\frac{54(+)54} {12}=9              

x=\frac{54(-)54} {12}=0          

therefore

The solutions are x=0 and x=9

4 0
3 years ago
Illustrative Example 1:
dlinn [17]

Given:

The data set is:

87, 84, 85, 85, 86, 90, 79, 82, 78, 76

To find:

The mode of the given data set.

Solution:

We have,

87, 84, 85, 85, 86, 90, 79, 82, 78, 76

Arrange the data values in the ascending order.

76, 78, 79, 82, 84, 85, 85, 86, 87, 90

We know that the mode of date set is the most frequency value of the data set.

From the above data set it is clear that the number 85 has the highest frequency 2.

Therefore, the mode of the data set is 85.

6 0
2 years ago
Prove that:
devlian [24]

Answer:

<u>Identities used:</u>

  • <em>1/cosθ = secθ</em>
  • <em>1/sinθ = cosecθ</em>
  • <em>sinθ/cosθ = tanθ</em>
  • <em>cosθ/sinθ = cotθ</em>
  • <em>sin²θ + cos²θ = 1</em>
<h3>Question 1 </h3>
  • (1 - sinθ)/(1 + sinθ) =        
  • (1 - sinθ)(1 - sinθ) / (1 - sinθ)(1 + sinθ) =
  • (1 - sinθ)² / (1 - sin²θ) =
  • (1 - sinθ)² / cos²θ

<u>Square root of it is:</u>

  • (1 - sinθ)/ cosθ =
  • 1/cosθ - sinθ / cosθ =
  • secθ - tanθ
<h3>Question 2 </h3>

<u>The first part without root:</u>

  • (1 + cosθ) / (1 - cosθ) =
  • (1 + cosθ)(1 + cosθ) / (1 - cosθ)(1 + cosθ)
  • (1 + cosθ)² / (1 - cos²θ) =
  • (1 + cosθ)² / sin²θ

<u>Its square root is:</u>

  • (1 + cosθ) / sinθ =
  • 1/sinθ + cosθ/sinθ =
  • cosecθ + cotθ

<u>The second part without root:</u>

  • (1 - cosθ) / (1 + cosθ) =
  • (1 - cosθ)²/ (1 + cosθ)(1 - cosθ) =
  • (1 - cosθ)²/ (1 - cos²θ) =
  • (1 - cosθ)²/sin²θ

<u>Its square root is:</u>

  • (1 - cosθ) / sinθ =
  • 1/sinθ - cosθ / sinθ =
  • cosecθ - cotθ

<u>Sum of the results:</u>

  • cosecθ + cotθ + cosecθ - cotθ =
  • 2cosecθ
4 0
3 years ago
Read 2 more answers
Did I get this correct
saveliy_v [14]

Answer:

Yep

Step-by-step explanation:

7 0
2 years ago
Use the formula A=bh , where A is the area, b is the base length, and h is the height of the parallelogram, to solve this proble
Anettt [7]
A=bh
1056=32<span>×b
</span>÷32  ÷32
3<span>3=h
The height is </span>3<span>3 inches.</span>
8 0
3 years ago
Read 2 more answers
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