Draw a rough sketch of path one on your graph

Answers

Answer 1

Answer: can you comment some details about your problem?

Answer 2

Answer: give more detail, like what is path one??

Related Questions

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What is the x coordinate?
Which permanent teeth eruption usually occurs first? A. Central incisors. B. Lateral incisors. C. Cuspids (canines). D. First premolars (Bicuspids).
If a || b, m<2 = 63°, and m<9 = 105°, give the measure of each angle.m<1 =M<3=m<4=m<5=m<6=m<7=m<8=m<10=m<11=m<12=m<13=m<14 =
Please help me with this im pretty confused ?

The midpoint of PQ is M(, –1). One endpoint is Q(3, –5). Which equations can be solved to determine the coordinates of P? Check all that apply.

Answers

Answer:

c and d. just did the lesson and its those two.

Step-by-step explanation:

Six more than three times a number is less than or equal to 96.

Answers

3x+6≤96
three x plus six is less than or equal to 96.

The value of a car has decreases by 5% each year. Sophie bought a car two years ago for £10000 work out the value now .

Answers

Answer:

£9000

Step-by-step explanation:

= £10 000×5×2

100

= 1000

= £10 000-1000

= £9000

Rowena has a $150,000 homeowner's insurance policy with a $1,000 deductible on her house. Her premium payment is $100 per month. If she makes a $5,000 claim on the policy, how much will she have to pay for her claim? A. $1,000 B. $1,200 C. $2,200 D. $4,000

Answers

She will have to pay her claim by
A. $1,000

Find three rational numbers equivalent to each of the following rational numbers. (i)−2/5 (ii)3/7

Answers

Step-by-step explanation:

(i)

-2/5 = -4/10 = -6/15 = -8/20

(ii)

3/7 = 6/14 = 9/21 = 12/28

we simply multiply top and bottom (numerator and denominator) by the same number.

Let a=x^2+4. Use a to find the solutions for the following equation: (x^2+4)^2+32=12x^2+48. Which one of the following are solutions for x? Select any/all that apply. -8, -2, 4, 0, 2, -4, 8

Answers

$(x^2+4)^2+32=12x^2+48 \n(x^2+4)^2+32=12(x^2+4) \ \ \ |-12(x^2+4) \n(x^2+4)^2-12(x^2+4)+32=0 \n\hbox{substitute a for } x^2+4: \na^2-12a+32=0 \na^2-4a-8a+32=0 \na(a-4)-8(a-4)=0 \n(a-8)(a-4)=0 \na-8=0 \ \lor \ a-4=0 \na=8 \ \lor \ a=4 \n \n\hbox{substitute 8 and 4 for a and solve for x:} \na=8 \n\Downarrow \n8=x^2+4 \ \ \ |-4 \n4=x^2 \nx=-2 \ \lor \ x=2 \n \na=4 \n\Downarrow \n4=x^2+4 \ \ \ |-4 \n0=x^2 \nx=0 \n \n\boxed{x=-2 \hbox{ or } x=0 \hbox{ or } x=2}$

The solutions for x are -2, 0, 2.

Answer:

-2,0,2

Step-by-step explanation:

The given equation is:

$(x^2+4)^(2)+32=12x^2+48$