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3 years ago

Define variable in mathematics

Mathematics

2 answers:

denis23 [38]3 years ago

8 0

Variable is the unknown value

Shtirlitz [24]3 years ago

6 0

Answer:

A letter to describe a unknown number in an equation

Step-by-step explanation:

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Y=6 X+2y=16 Can someone one help me out its a substitution and system of equations graph

Nikitich [7]

If you know that y = 6. You just have to switch Y by 6.

x + 2y = 16
x + 2*6 = 16
x + 12 = 16
x = 16 - 12
x= 4

4 + 2*6 = 16

Hope this helps !

Photon

7 0

4 years ago

In a class of 25 students, 80% attended a field trip. How many students went on the field trip?

Dmitry_Shevchenko [17]

Answer:

20 Students :)

Step-by-step explanation:

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3 years ago

Read 2 more answers

If the police car accelerates uniformly at 2.00 m/s2 and overtakes the speeder after accelerating for 9.00 s , what was the spee

Vika [28.1K]

2.00m/s²×9.00s=18m/s
a×t=speed

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3 years ago

Find the sum. 7/30 + 9/20 PLS

e-lub [12.9K]

Answer:

41/60

Step-by-step explanation:

Convert the fractions into like denominators: 7/30 = 140/600, 9/20 = 270/600

Add: 140/600+270/600 = 410/600

Divide into a smaller fraction: 41/60

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3 years ago

Let a = x2 + 4. Use a to find the solutions for the following equation:

Zepler [3.9K]

The solutions for x are -2, 0, 2

Solution:

Given that,

$\text { Let } a = x^2 + 4$

Given equation is:

$(x^2 + 4)^2 + 32 = 12x^2 + 48$

$(x^2 + 4)^2 + 32 = 12(x^2 + 4)$

$\text { Subtsitute } a = x^2 + 4 \text{ in above equation }$

$a^2 + 32 = 12a\\\\a^2 -12a + 32 = 0$

$\text{ Let us factorize the above equation }\\\\\text{ Splitting the middle term -12a as -4a - 8a we get, }$

$a^2 -4a - 8a + 32 = 0$

Taking "a" as common term from first two terms and taking "-8" as common from last two terms

$a(a-4)-8(a - 4) = 0$

$\text{Taking (a - 4) as common term, }\\\\(a - 4)(a - 8) = 0$

$\text{Equating to zero we get, }\\\\(a - 4) = 0 \text{ or } (a - 8) = 0\\\\a = 4 \text{ or } a = 8$

$\text{Now substitute the value of a = 8 and a = 4 in: }\\\\a = x^2 + 4$

$\text{ For a = 8: }\\\\a = x^2 + 4\\\\8 = x^2 + 4\\\\x^2 = 4\\\\x = \pm2\\\\x = +2 \text{ or } -2$

$\text{For a = 4: }\\\\a = x^2 + 4\\\\4 = x^2 + 4\\\\x = 0$

$\text{Thus solutions of } x \text{ are: }\\\\x = 0 \text{ or } x = 2 \text{ or } x = -2$

8 0

3 years ago