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3 years ago

8

Which system of equations does not have a real solution?

2 answers:

dolphi86 [110]3 years ago

7 0

Answer:

It's A.

Step-by-step explanation:

Let's look at option A:

From the second equation y = -10 - x. Substituting in the first equation:

-10 - x = x^2 + 3x - 5

x^2 + 4x + 5 = 0

Checking the discriminant b^2 - 4ac we get 16 - 4*1*5 = -4 so there are no real roots. (A negative discriminant means no real roots).

So A has no real solution.

B.

x^2 + 3x - 5 = (20 - 4x)/5 = 4 - 0.8x

x^2 +3.8x - 9 = 0

b^2 - 4ac = (3.8)^2 - 4*1*-9 = 50.44 (positive) so there are real roots.

C.

x^2 + 3x - 5 = -9 - x

x^2 + 4x + 4 = 0

b^2 - 4ac = 4^2 - 4*1*4 = 0 so there are real roots.

Tatiana [17]3 years ago

6 0

Answer:

Correct answer is A!

Step-by-step explanation:

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2 years ago

The price of an item has risen to $ 248 today. Yesterday it was $ 160 . Find the percentage increase.

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4 0

3 years ago

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Sati [7]

Answer:

-89

Step-by-step explanation:

-89

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5 0

3 years ago

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Write the equation of a line that passes through (2,6) and (-2,2)

Viktor [21]

Answer:

y = x + 4

Step-by-step explanation:

The equation of the line is

y = mx + c

Step 1: find the slope

m = y2 - y1 / x2 - x1

Give points

( 2 , 6) ( -2 ,2)

x1 = 2

y1 = 6

x2 = -2

y2 = 2

m = 2 - 6 / -2 - 2

m = -4 / -4

m = 1

y = mx + c

y = 1x + c

y = x + c

Step 2: sub any of the two points given into the equation

( 2 , 6)

x = 2

y = 6

y = x + c

6 = 2 + c

6 - 2 = c

c = 4

Step 3: sub c into the equation

y = x + 4

The equation of the line is

y = x + 4

5 0

3 years ago

Which pair of funtions is not a pair of inverse functions? please help!!

antiseptic1488 [7]

Answer:

$f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Step-by-step explanation:

we know that

To find the inverse of a function, exchange variables x for y and y for x. Then clear the y-variable to get the inverse function.

we will proceed to verify each case to determine the solution of the problem

<u>case A)</u> $f(x)=\frac{x+1}{6} , g(x)=6x-1$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y+1}{6}$

Isolate the variable y

$6x=y+1$

$y=6x-1$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=6x-1$

therefore

f(x) and g(x) are inverse functions

<u>case B)</u> $f(x)=\frac{x-4}{19} , g(x)=19x+4$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y-4}{19}$

Isolate the variable y

$19x=y-4$

$y=19x+4$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=19x+4$

therefore

f(x) and g(x) are inverse functions

<u>case C)</u> $f(x)=x^{5}, g(x)=\sqrt[5]{x}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=y^{5}$

Isolate the variable y

fifth root both members

$y=\sqrt[5]{x}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\sqrt[5]{x}$

therefore

f(x) and g(x) are inverse functions

<u>case D)</u> $f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}$

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

$x=\frac{y}{y+20}$

Isolate the variable y

$x(y+20)=y$

$xy+20x=y$

$y-xy=20x$

$y(1-x)=20x$

$y=20x/(1-x)$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=20x/(1-x)$

$\frac{20x}{1-x}\neq \frac{20x}{x-1}$

therefore

f(x) and g(x) is not a pair of inverse functions

7 0

3 years ago

Read 2 more answers