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3 years ago

Solve for x & y: x^2+y^2=20. 3x+y=2

2 answers:

6 0

You're solving a system here, easy peasy. Solve the second equation for y to get y = 2 - 3x. Now in the first equation, sub in this y value where you see a y and now the equation is in terms of x only: x^2 + (2-3x)^2 = 20. You have to FOIL out the (2-3x)^2 first to get x^2 + 4 - 12x + 9x^2 = 20. Now combine like terms to get 10x^2 - 12x + 4 = 20. Move the 20 over by subtraction to get
10x^2 - 12x - 16 = 0. Factor this now, taking a 2 out first: 2(5x^2 - 6x -8)=0 and you get factors of 2(x-2)(5x+4)=0. x=2 or x = -4/5. If x = 2, then y = -4. If x = -4/5, then y = 22/5

Lerok [7]3 years ago

5 0

Answer:

Step-by-step explanation:

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What should the graph of two lines look like when there are no solutions that they have in common?

nataly862011 [7]

If the graphs of the equations do not intersect

4 0

1 year ago

Read 2 more answers

If there are 6 serving in a 2/3 pound lb package of pound is in each serving

Serhud [2]

Answer:

$\boxed{\math{\frac{1}{9}\text{ lb}}}$

Step-by-step explanation:

$\text{Size of one serving} = \dfrac{\text{Total size}}{\text{No of servings}} = \dfrac{\frac{2}{3}\text{ lb}}{\text{6 servings}}\\\\\text{Change the 6 to a fraction and change divide to multiply}\\\\\dfrac{2}{3} \div 6 = \dfrac{2}{3} \times \dfrac{1}{6}$

$\text{Cancel the 2s}\\\\\dfrac{2}{3} \times \dfrac{1}{6} = \dfrac{1}{3} \times \dfrac{1}{3}\\\\\text{Multiply numerators and denominators}\\\\\dfrac{1}{3} \times \dfrac{1}{3} = \dfrac{1}{9}\\\\\text{Each serving contains }\boxed{\mathbf{\frac{1}{9}\textbf{ lb}}}$

3 0

2 years ago

This table models a linear function. x −8 −4 4 8 y −3 0 6 9 Enter the coordinates of the y-intercept in the boxes.

adell [148]

Answer:

y-intercept is 3

Step-by-step explanation:

We are given table

x: -8 -4 4 8

y: -3 0 6 9

We can select any two points

$x_1=-8,y_1=-3$

$x_2=-4,y_2=0$

Firstly, we will find slope

$m=\frac{y_2-y_1}{x_2-x_1}$

now, we can plug values

$m=\frac{0+3}{-4+8}$

$m=\frac{3}{4}\quad$

now, we can use point slope form of line

$y-y_1=m(x-x_1)$

we can plug values

$y+3=\frac{3}{4}(x+8)$

now, we can solve for y

$y=\frac{3}{4}x+3$

For finding y-intercept , we can plug x=0

$y=\frac{3}{4}\times 0+3$

$y=0+3$

$y=3$

So,

y-intercept is 3

8 0

3 years ago

A house and a lot is appraised at $212,400. The value of the house is five times more the value of the lot. What is the house wo

Anuta_ua [19.1K]

The house is worth $177,000
the lot is worth $35,400

5 0

3 years ago

If y varies directly as (x+3) and inversely as (x-3) and y=21 when x=4, what is the equation for y in terms of x?

igor_vitrenko [27]

$\bf \qquad \qquad \textit{double proportional variation}
\\\\
\begin{array}{llll}
\textit{\underline{y} varies directly with \underline{x}}\\
\textit{and inversely with \underline{z}}
\end{array}\implies y=\cfrac{kx}{z}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\$

$\bf \textit{\underline{y} varies directly as (x+3) and inversely as (x-3)}\implies y=\cfrac{k(x+3)}{x-3}
\\\\\\
\textit{we also know that }
\begin{cases}
y=21\\
x=4
\end{cases}\implies 21=\cfrac{k(4+3)}{4-3}\implies 21=\cfrac{k(7)}{1}
\\\\\\
\cfrac{21}{7}=k\implies 3=k\qquad thus\qquad \qquad \boxed{y=\cfrac{3(x+3)}{x-3}}$

6 0

3 years ago