All categories

3 years ago

Watch ordered pair is not a point on the graph of y= -5x + 2?

Mathematics

1 answer:

-BARSIC- [3]3 years ago

4 0

D (2,-8) is not a point on the graph

You might be interested in

Find y, at and tu I need help please help

lesantik [10]

Answer:

there is no picture... but i will try my best to help you if you post picture in another question

8 0

3 years ago

Read 2 more answers

Can someone help me with this i need it for today

N76 [4]

Answer: 2y=-6.4 = y=−3.2

-4y= -3.2 = y=0.8

Step-by-step explanation:

only answering what i can sorry

3 0

3 years ago

Best explained and correct answer gets brainliest.

ELEN [110]

3275 = x * (1+0.075/12)^(12*10)

3275 = x * (2.11206463)

x= 3275/2.11206463 = 1550.6154

x=1550.62

she needs to deposit $1,550.62

8 0

3 years ago

HELPLPLPPLPPLPPLPPLPLPLPL

kipiarov [429]

Answer:

Bro the answer is LITERALLY 1

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Y=-x^2+2x+10 y=x+2 Substitution Please show your work Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

Find the values of y

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

3 years ago