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bija089 [108]
3 years ago
8

Which ordered pair is the solution of the system?

Mathematics
2 answers:
lubasha [3.4K]3 years ago
7 0
Solution is where the graphs intercepts. Here is only one point they are touching,
(1,1) is a solution.

kolezko [41]3 years ago
4 0
(1,1) would be the answer

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Someone plz help me!!
marta [7]

Answer:

P'=(0,3)

Step-by-step explanation:

P'=(-5+5, 7-4)

8 0
2 years ago
The graph of the equation y=|x+5| is shifted 2 units up and 6 units to the right. Rewrite the equation of the new graph for the
DerKrebs [107]

Answer:

Step-by-step explanation:

To move the graph 6 units to the right, subtract 6 in the absolute value.

y=|x+5-6|

To move it up, add 2 outside the absolute value.

y=|x+5-6|+2

Now remove the absolute values since x>1.

y=x+5-6+2

y=x+1

4 0
3 years ago
Lim (sin(pi/x)) as x tends to 0 from both sides​
strojnjashka [21]

Hello from MrBillDoesMath!

Answer:

Limit does not exist.

Discussion:

The function 1/x has a vertical asymptote as x approaches 0 so

sin (pi/x) has no limit as x approaches 0. In fact, it oscillates wildly between -1 and 1 as x approaches 0. See attached graph of function

Thank you,

MrB

8 0
3 years ago
Suggest five questions for a questionnaire to discover what opinions people in class have about school.
Snezhnost [94]

Answer:

Which activities in the classroom do you enjoy the most? ...

Given a chance, what is one change that you would like to see? ...

Do you have supportive classmates? ...

What motivates you to learn more? ...

Do you think that the school provides you with adequate sports facilities?

How much time do you spend on homework every night?

Which classroom activities do you learn from the most?

What are three things that can improve the class most?

8 0
2 years ago
1.) What is the equation of the path of firework #1? Write your equation in general form.
valina [46]

1. I'm assumig that the paths are perfect parabolas

this means that their general forms can be written in y=ax^2+bx+c

it's easier to find vertex form first then expand to get general form

vertex form is y=a(x-h)^2+k where the vertex is (h,k) and a is a constant


firework #1

vertex is (10,50), so (h,k)=(10,50) and h=10, k=50

h_1=a(t-10)^2+50

to find the value of a, subsitute another point

(0,0)

0=a(0-10)^2+50)

0=100a+50

a=\frac{-1}{2}

so the equation in vertex form is h_1=\frac{-1}{2}(t-10)^2+50

expand to get general form

h_1=\frac{-1}{2}(t^2-20t+100)+50

h_1=\frac{-1}{2}t^2+10t-50+50

h_1=\frac{-1}{2}t^2+10t





2.

same as last time

vertex is (10,72) so (h,k)=(10,72) so h=10 and k=72

equation is h_2=a(t-10)^2+72

find a

use another point

(0,22)

22=a(0-10)^2+72

22=100a+72

-50=100a

a=\frac{-1}{2}

so the equation in vertex form is h_2=\frac{-1}{2}(t-10)^2+72




3.

range is the numbers that h is allowed to be

think about what h represents. it represents the height of the rocket

from the graph, we can see that the lowest possible height is 0yd and the highest height is 50yd

so range is 0 to 50 or 0≤h≤50


domain is the numbers that t is allowed to be

think about what t represents. it represents how long the rocket has been flying

it will stop flying when it hits the ground or at t=20

it starts flying at t=0

so domain is from 0 to 20 or 0≤t≤20

3 0
3 years ago
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