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3 years ago

How to solve 11x-13y=89 and -11x+13y=107 by substitution

1 answer:

8 0

$\left\{\begin{array}{ccc}11x-13y=89&/+13y\\-11x+13y=107&/\cdot(-1)\end{array}\right\\\\\left\{\begin{array}{ccc}11x=13y+89\\11x-13y=-107\end{array}\right\\\\substitute:\\\\(13y+89)-13y=107\\13y-13y+89=107\\89=107-FLASE\\\\Answer:No\ solution\ (x;\ y\in\O)$

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If f(x) = 2x2 + 1 and g(x) = x2 – 7, find (f – g)(x).

nadya68 [22]

Answer: The required value is

$(f-g)(x)=x^2+8.$

Step-by-step explanation: The given functions are:

$f(x)=2x^2+1,\\\\g(x)=x^2-7.$

We are given to find the value of $(f-g)(x).$

We know that, if s(x) and t(x) are any two functions of a variable x, then we have

$(s-t)(x)=s(x)-t(x).$

Therefore, we have

$(f-g)(x)\\\\=f(x)-g(x)\\\\=(2x^2+1)-(x^2-7)\\\\=2x^2+1-x^2+7\\\\=x^2+8.$

Thus, the required value is

$(f-g)(x)=x^2+8.$

5 0

2 years ago

A. f(x) = l-2xl - 3 Domain: {1, 2, 3} Find the range

gizmo_the_mogwai [7]

Hello there!

We are given the function:

$\displaystyle \large{ f(x) = | - 2x| - 3} \\$

To find the range, we know that domain is the set of all x-values and also called 'input' while range is the set of all y-values and also called 'output'.

Basic Function - you add the input, you get the output. You add x-value in a function, you get y-value. You add domain, you get range.

So, we substitute x = 1,2 and 3 in the function.

x = 1

$\displaystyle \large{ f(1) = | - 2(1)| - 3} \\ \displaystyle \large{ f(1) = | - 2| - 3} \\$

Recall that any numbers in absolute value are always positive.

$\displaystyle \large{ f(1) = 2 - 3} \\ \displaystyle \large{ f(1) = - 1} \\$

x = 2

$\displaystyle \large{ f(2) = | - 2(2)| - 3} \\ \displaystyle \large{ f(2) = | - 4 | - 3} \\ \displaystyle \large{ f(2) = 4 - 3} \\ \displaystyle \large{ f(2) = 1}$

x = 3

$\displaystyle \large{ f(3) = | - 2(3)| - 3} \\ \displaystyle \large{ f(3) = | - 6| - 3} \\ \displaystyle \large{ f(3) = 6 - 3} \\ \displaystyle \large{ f(3) = 3}$

Therefore, Range: {-1,1,3}

Let me know if you have any questions!

Topic: Absolute Value Function / Modulus Function

6 0

2 years ago

the population of cities A and B are 3.8 x 10^5 and 2,160,000 , respectively . the population of city C is twice the population

erma4kov [3.2K]

The answer is 4, 320, 000

explanation: the problem is asking what the population of city c is, and to do that you need to multiply city b times 2. therefore you do not need the population of city a, it is just there to confuse you

5 0

2 years ago

PLEASE HELP ASAP!!!!!!!!!!!

marusya05 [52]

Answer:(1) The correct option is b. (2) The correct option is c.

Explanation:

(1)

It is given that the △ABD≅△FEC.

So by (CPCTC) corresponding parts of congruent triangles are congruent.

$\angle CFE=\angle DAB$