All categories

2 years ago

Systems of equations with graphing

Mathematics

2 answers:

frez [133]2 years ago

5 0

The points intercept at x= -7/5 and y=8/5

OlgaM077 [116]2 years ago

3 0

Answer:

x= negative 1 and 2/5 y=1 and 3/5

Step-by-step explanation:

1. Start by putting both equations into slope-intercept form. (y=mx+b)

-3x+3y=9

-3x+3x+3y=9+3x

3y=9+3x

3y/3=9+3x/3

y=3+x

2x-7y=-14

2x-2x-7y=-14-2x

-7y=-14-2x

-7y/-7=-14/-7-2x/-7

y=2+2/7x

2. Graph the equations and find where they intercept.

You might be interested in

Solve for x.<br> 5x−2(x+1)=14 <br><br> x =

IceJOKER [234]

Answer:

x=16/3

Step-by-step explanation:

5x−2(x+1)=14

Distribute

5x -2x -2 =14

Combine like terms

3x -2 =14

Add 2 to each side

3x-2+2 =14+2

3x=16

Divide each side by 3

3x/3 = 16/3

x=16/3

8 0

3 years ago

Read 2 more answers

If the area of square is 2401 sqaure metre,find the perimeter of the sqaure

Tcecarenko [31]

Answer

P=196

P=4A=4·2401=196

6 0

3 years ago

Read 2 more answers

*Pls help me*<br> No links

harkovskaia [24]

Answer:

See below ↓

Step-by-step explanation:

a. 4 ÷ (2/3) = 12/3 ÷ 2/3 = 12 ÷ 2 = 6

b. 4 ÷ 1 = 4

Attachment below.

6 0

2 years ago

Read 2 more answers

Graph the parabola y=-3x^2+24x-42

Usimov [2.4K]

Based on the vertex and table this is the graph.

5 0

3 years ago

Find f. f ″(x) = x^−2, x > 0, f(1) = 0, f(6) = 0

marin [14]

If you do in fact mean $f(1)=f(6)=0$ (as opposed to one of these being the derivative of $f$ at some point), then integrating twice gives

$f''(x) = -\dfrac1{x^2}$

$f'(x) = \displaystyle -\int \frac{dx}{x^2} = \frac1x + C_1$

$f(x) = \displaystyle \int \left(\frac1x + C_1\right) \, dx = \ln|x| + C_1x + C_2$

From the initial conditions, we find

$f(1) = \ln|1| + C_1 + C_2 = 0 \implies C_1 + C_2 = 0$

$f(6) = \ln|6| + 6C_1 + C_2 = 0 \implies 6C_1 + C_2 = -\ln(6)$

Eliminating $C_2$ , we get

$(C_1 + C_2) - (6C_1 + C_2) = 0 - (-\ln(6))$

$-5C_1 = \ln(6)$

$C_1 = -\dfrac{\ln(6)}5 = -\ln\left(\sqrt[5]{6}\right) \implies C_2 = \ln\left(\sqrt[5]{6}\right)$

Then

$\boxed{f(x) = \ln|x| - \ln\left(\sqrt[5]{6}\right)\,x + \ln\left(\sqrt[5]{6}\right)}$

3 0

2 years ago