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Marrrta [24]
3 years ago
14

Uhh... help with numb er seven because brother can’t solve it

Mathematics
1 answer:
Akimi4 [234]3 years ago
7 0
It’s a parabola which is y= x^2. We know this because it’s vertex hits the origin. There are 2 answers there for it so just pick one.
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Can someone please help me with this its urgent I don’t get it
Varvara68 [4.7K]

Answer:

$1,221

Step-by-step explanation:

0.11×$1,100=$121

$121+ $1100= $1221

4 0
3 years ago
Select ALL expressions that have a value less than 9.
GrogVix [38]

Answer:

What is this?

Can't understand anything!

6 0
2 years ago
Read 2 more answers
What are the solutions of the system?
siniylev [52]
y=x^2-3x+2 \\y=4x-4

A. (1, 0) and (–6, –28) 

B. (1, 20) and (6, 0) 

C. (1, 0) and (6, 20) 

D. no solution


Correct answer is C.


y=x^2-3x+2 \\y=4x-4\\ \\4x-4=x^2-3x+2 \\x^2-7x+6=0 \\x^2-6x-x+6=0 \\x(x-6)-(x-6)=0 \\(x-6)(x-1)=0 \\x=6\,\text{ or }x=1 \\y=4\times 6-4=20 \,\text{ or }y=4\times 1-4=0 \\ \\(6,20)\,\text{ and }(1,0)

8 0
3 years ago
Read 2 more answers
In quadratic drag problem, the deceleration is proportional to the square of velocity
Mars2501 [29]
Part A

Given that a= \frac{dv}{dt} =-kv^2

Then, 

\int dv= -kv^2\int dt \\  \\ \Rightarrow v(t)=-kv^2t+c

For v(0)=v_0, then

v(0)=-kv^2(0)+c=v_0 \\  \\ \Rightarrow c=v_0

Thus, v(t)=-kv(t)^2t+v_0

For v(t)= \frac{1}{2} v_0, we have

\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\  \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\  \\ \Rightarrow kv_0t=2 \\  \\ \Rightarrow t= \frac{2}{kv_0}


Part B

Recall that from part A, 

v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\  \\ \Rightarrow dx=-kv^2tdt+v_0dt \\  \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\  \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a

Now, at initial position, t = 0 and v=v_0, thus we have

x=a

and when the velocity drops to half its value, v= \frac{1}{2} v_0 and t= \frac{2}{kv_0}

Thus,

x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\  \\ =- \frac{1}{2k} + \frac{2}{k} +a

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

- \frac{1}{2k} + \frac{2}{k} +a-a \\  \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}
7 0
3 years ago
What is the domain of the function on the graph?
Dennis_Churaev [7]

Answer:

All real numbers greater than or equal to -3

Step-by-step explanation:

First look at graph where the line points to which direction of the graph

And look for any closed or open circles in the graph

Since in the graph has a close circle at (-3,-2) meaning it includes that x-value for its domain.

With the graph going to positive infinity it states that the domain is all real numbers.

So in conclusion it has a domain of all real numbers greater than or equal to -3

5 0
4 years ago
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