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3 years ago

I need this ctugkvhjkibnlbk;vhiglcufxoryifcuglvihk;b j;hviglucforxiy6eryfcuglvhk

Mathematics

2 answers:

LUCKY_DIMON [66]3 years ago

7 0

Answer:

11 * 7 + 11 * 4

Step-by-step explanation:

given 11(7+4)

we need to use distributive property

To do so we multiply 11 by whats inside of the parenthesis

so 11(7+4) is equivalent to 11 * 7 + 11 * 4

Orlov [11]3 years ago

3 0

Answer:

11(7 + 4) = 11 × 7 + 11 × 4

Step-by-step explanation:

11(7 + 4) = _ x 7 + 11 x _

use distributive property:

11(7 + 4) = 7 x 11 + 7 x 11

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Which expression has a value of 20?

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Stephanie grew a total of 70 pounds of tomatoes to sell. She plans to sell the tomatoes for $3.50 per pound. The equation below

Alexxandr [17]

Answer:

If Stephanie sold the 70 pounds of tomatoes, her profit would be 135 dollars.

Step-by-step explanation:

You know that the equation f(x) = 3.50*x - 110 represents the profit in dollars Stephanie will earn by selling x pounds of the tomatoes.

If Stephanie sold the 70 pounds of tomatoes, you want to know what the profit would be. So to calculate the profit you simply replace in f (x), the x pounds of tomatoes by 70.

f(70)=3.50*70 - 110

Solving:

f(70)= 245 - 110

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<u><em>If Stephanie sold the 70 pounds of tomatoes, her profit would be 135 dollars.</em></u>

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3 years ago

Definition of Properties and attributes of functions

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3 years ago

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Solve the following pairs of equations for x and y:

Delicious77 [7]

Answer:

<h3><u>Given </u><u>Question:</u><u>-</u></h3>

Solve the following pair of equations for x and y :

$\rm :\longmapsto\:\dfrac{ {a}^{2} }{x} - \dfrac{ {b}^{2} }{y} = 0, \: \: \dfrac{ {a}^{2}b}{x} + \dfrac{ {b}^{2} a}{y} = a + b$

$\red{\large\underline{\sf{Solution-}}}$

Given pair of equations are

$\rm :\longmapsto\:\dfrac{ {a}^{2} }{x} - \dfrac{ {b}^{2} }{y} = 0 - - - - (1)$

and

$\rm :\longmapsto\:\dfrac{ {a}^{2} b}{x} + \dfrac{ {b}^{2}a }{y} = a + b - - - - (2)$

On multiply equation (1) by a, we get

$\rm :\longmapsto\:\dfrac{ {a}^{3} }{x} - \dfrac{ {b}^{2} a}{y} = 0 - - - (3)$

On adding equation (3) and (2), we get

$\rm :\longmapsto\:\dfrac{ {a}^{3} }{x} + \dfrac{ {a}^{2} b}{x} = a + b$

$\rm :\longmapsto\:\dfrac{ {a}^{3} + {a}^{2} b}{x} = a + b$

$\rm :\longmapsto\:\dfrac{{a}^{2}(a + b)}{x} = a + b$

$\\ \rm\implies \:\boxed{\tt{ \: \: x \: \: = \: \: {a}^{2} \: \: }} \\$

On substituting the value of x in equation (1), we get

$\rm :\longmapsto\:\dfrac{ {a}^{2} }{ {a}^{2} } - \dfrac{ {b}^{2} }{y} = 0$

$\rm :\longmapsto\:1 - \dfrac{ {b}^{2} }{y} = 0$

$\rm :\longmapsto\: - \dfrac{ {b}^{2} }{y} = - 1$

$\rm :\longmapsto\: \dfrac{ {b}^{2} }{y} = 1$

$\\ \rm\implies \:\boxed{\tt{ \: \: y \: \: = \: \: {b}^{2} \: \: }} \\$

<h3><u>VERIFICATION:</u></h3>

Consider the first equation

$\rm :\longmapsto\:\dfrac{ {a}^{2} }{x} - \dfrac{ {b}^{2} }{y} = 0$

On substituting the values of x and y, we get

$\rm :\longmapsto\:\dfrac{ {a}^{2} }{ {a}^{2} } - \dfrac{ {b}^{2} }{ {b}^{2} } = 0$

$\rm :\longmapsto\: 1 - 1 = 0$

$\rm\implies \:0 = 0$

<u>Hence, Verified </u>

So, Solution of pair of equations is

$\purple{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\bf{x \: = \: {a}^{2} } \\ \\ &\bf{y \: = \: {b}^{2} } \end{cases}\end{gathered}\end{gathered}}$

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3 years ago