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2 years ago

Help please =] Solve system of equations using any method 2x-6y=8 and 2x-6y=3

2 answers:

6 0

Let z be any number (like x would be)

if we set z equal to 2x-6y, then we get z = 2x-6y

The system

2x-6y = 8
2x-6 = 3

turns into this new system

z = 8
z = 3

but z is a single number. It can't be both 8 and 3 at the same time. So there are no solutions

romanna [79]2 years ago

3 0

The answer is no solution. If you use elimination, by subtracting the second equation from the first, you come out with 0=5, which is a false statement. Therefore, there is no solution.

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May I get some help? Please and thank you

katrin [286]

Answer:V x 80 = 36 Original value is $0.45

Step-by-step explanation: V is the original value and your x it by 80

4 0

2 years ago

The product of two numbers is 240. The first number is 8 less than the second number. Write an equation can be used to find x

DaniilM [7]

Answer:

The equation x = y-8

The equation (y-8) y = 240

The first number x = 12 and second number y =20

Step-by-step explanation:

Explanation:-

Step(l):-

Let 'x' be the first number and 'y' be the second number

Given the product of two numbers is 240

Given data x y = 240 …(l)

Given data the first number is 8 less than the second number.

x = y-8...(ll)

Substitute (ll) in equation (l) , we get

(y-8) y = 240

$y^{2} -8y -240 =0$

$y^{2} -20y+12y -240 =0$

$y(y -20)+12(y -20) =0$

$(y +12)(y -20) =0$

y = -12 and y = 20

Y= -12 is not satisfied

we can choose y = 20

Step(ll):-

Given data x =y-8

substitute value y = 20 in x = y -8

x = 20-8 = 12

Final answer:-

The first number x = 12 and second number y =20

4 0

2 years ago

Item 7

Mariulka [41]

Answer:

$A = 74.7^\circ$

$B = 42.5^\circ$

$C = 62.8^\circ$

Step-by-step explanation:

Given

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

For AB

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$d = \sqrt{(-1 - 2)^2 + (2 - 8)^2$

$d = \sqrt{(-3)^2 + (-6)^2$

$d = \sqrt{45$

So:

$AB = \sqrt{45$

For BC

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

$BC = \sqrt{(2 - 4)^2 + (8 - 1)^2$

$BC = \sqrt{(-2)^2 + (7)^2$

$BC = \sqrt{53$

For AC

$A = (-1,2) \to (x_1,y_1)$

$C = (4,1) \to (x_3,y_3)$

$AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2$

$AC = \sqrt{(-5)^2 + (1)^2$

$AC = \sqrt{26$

So, we have:

$AB = \sqrt{45$

$BC = \sqrt{53$

$AC = \sqrt{26$

By representation

$AB \to c$

$BC \to a$

$AC \to b$

So, we have:

$a = \sqrt{53$

$b = \sqrt{26$

$c = \sqrt{45$

By cosine laws, the angles are calculated using:

$a^2 = b^2 + c^2 -2bc \cos A$

$b^2 = a^2 + c^2 -2ac \cos B$

$c^2 = a^2 + b^2 -2ab\ cos C$

$a^2 = b^2 + c^2 -2bc \cos A$

$(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A$

$53 = 26 +45 - 2 * 34.21 * \cos A$

$53 = 26 +45 - 68.42 * \cos A$

Collect like terms

$53 - 26 -45 = - 68.42 * \cos A$

$-18 = - 68.42 * \cos A$

Solve for $\cos A$

$\cos A =\frac{-18}{-68.42}$

$\cos A =0.2631$

Take arc cos of both sides

$A =\cos^{-1}(0.2631)$

$A = 74.7^\circ$

$b^2 = a^2 + c^2 -2ac \cos B$

$(\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B$

$26 = 53 +45 -97.67 * \cos B$

Collect like terms

$26 - 53 -45= -97.67 * \cos B$

$-72= -97.67 * \cos B$

Solve for $\cos B$

$\cos B = \frac{-72}{-97.67}$

$\cos B = 0.7372$

Take arc cos of both sides

$B = \cos^{-1}(0.7372)$

$B = 42.5^\circ$

For the third angle, we use:

$A + B + C = 180$ --- angles in a triangle

Make C the subject

$C = 180 - A -B$

$C = 180 - 74.7 -42.5$

$C = 62.8^\circ$

8 0

1 year ago

HELP with these questions

zlopas [31]

Step-by-step explanation:

transform the parent graph of f(x) = ln x into f(x) = - ln (x - 4) by shifting the parent graph 4 units to the right and reflecting over the x-axis

(???, 0): 0 = - ln (x - 4)

$\frac{0}{-1} = \frac{-ln (x - 4)}{-1}$

0 = ln (x - 4)

$e^{0} = e^{ln (x - 4)}$

1 = x - 4

+4 +4

5 = x

(5, 0)

(???, 1): 1 = - ln (x - 4)

$\frac{0}{-1} = \frac{-ln (x - 4)}{-1}$

1 = ln (x - 4)

$e^{1} = e^{ln (x - 4)}$

e = x - 4

+4 +4

e + 4 = x

6.72 = x

(6.72, 1)

Domain: x - 4 > 0

+4 +4

x > 4

(4, ∞)

Vertical asymptotes: there are no vertical asymptotes for the parent function and the transformation did not alter that

No vertical asymptotes

*************************************************************************

transform the parent graph of f(x) = 3ˣ into f(x) = - 3ˣ⁺⁵ by shifting the parent graph 5 units to the left and reflecting over the x-axis

Domain: there is no restriction on x so domain is all real number

(-∞, ∞)

Range: there is a horizontal asymptote for the parent graph of y = 0 with range of y > 0. the transformation is a reflection over the x-axis so the horizontal asymptote is the same (y = 0) but the range changed to y < 0.

(-∞, 0)

Y-intercept is when x = 0:

f(x) = - 3ˣ⁺⁵

= - 3⁰⁺⁵

= - 3⁵

= -243

Horizontal Asymptote: y = 0 (explanation above)

5 0

2 years ago

What is the center and radius of the circle? * x² – 2x + 5+ y² = 8

wolverine [178]

Answer: center (1;0), radius = 2

Step-by-step explanation:

x² – 2x + 5+ y² = 8

x² – 2x + 5+ y² = x² – 2x + 1 + 4 + y² = (х-1)² +4 + y²

(х-1)² +4 + y² = 8

(х-1)² + y² = 8 - 4 = 4

(х-1)² + y² = 2²

center (1;0), radius = 2

5 0

2 years ago