Priya starts with $50 in her bank account. She then deposits $20 each week for 12 weeks. Write an equation that represents the relationship between the dollar amount (y) in her bank account and the number of weeks (x) of savings

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Answer 1

Answer:

Answer: y=mx+b is the proper equation

Step-by-step explanation:

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1) Solve by using the perfect squares method. x2 + 8x + 16 = 0 2) Solve. x2 – 5x – 6 = 0

3) What value should be added to the expression to create a perfect square? x2 – 20x

4) Solve. x2 + 8x – 8 = 0

5) Solve: 2x2 + 12x = 0

6) Solve each problem by using the quadratic formula. Write solutions in simplest radical form. 2x2 – 2x – 1 = 0

7) Calculate the discriminant. x2 – x + 2 = 0

8) Calculate the discriminant and use it to determine how many real-number roots the equation has. 3x2 – 6x + 1 = 0

9) Without drawing the graph of the equation, determine how many points the parabola has in common with the x-axis and whether its vertex lies above, on, or below the x-axis. y = 2x2 + x – 3

10) Without drawing the graph of the equation, determine how many points the parabola has in common with the x-axis and whether its vertex lies above, on, or below the x-axis. y = x2 – 12x + 12

Answers

1)
$x^2+8x+16=0 \n(x+4)^2=0 \nx+4=0 \n\boxed{x=-4}$

2)
$x^2-5x-6=0 \nx^2-6x+x-6=0 \nx(x-6)+1(x-6)=0 \n(x+1)(x-6)=0 \nx+1=0 \ \lor \ x-6=0 \nx=-1 \ \lor \ x=6 \n\boxed{x=-1 \hbox{ or } x=6}$

3)
$\hbox{a perfect square:} \n (x-a)^2=x^2-2xa+a^2 \n \n 2xa=20x \n a=(20x)/(2x) \n a=10 \n \n a^2=10^2=100 \n \n \hbox{the expression:} \n x^2-20x+100 \n \n \boxed{\hbox{100 should be added to the expression}}$

4)
$x^2+8x-8=0 \n \na=1 \n b=8 \n c=-8 \n \Delta=b^2-4ac=8^2-4 * 1 * (-8)=64+32=96 \n√(\Delta)=√(96)=√(16 *6)=4√(6) \n \nx=(-b \pm √(\Delta))/(2a)=(-8 \pm 4√(6))/(2 * 1)=(2(-4 \pm 2√(6)))/(2)=-4 \pm 2√(6) \n\boxed{x=-4-2√(6) \hbox{ or } x=-4+2√(6)}$

5)
$2x^2+12x=0 \n2x(x+6)=0 \n2x=0 \ \lor \ x+6=0 \nx=0 \ \lor \ x=-6 \n\boxed{x=-6 \hbox{ or } x=0}$

6)
$2x^2-2x-1=0 \n \na=2 \n b=-2 \n c=-1 \n \Delta=b^2-4ac=(-2)^2-4 * 2 * (-1)=4+8=12 \n√(\Delta)=√(12)=√(4 * 3)=2√(3) \n \nx=(-b \pm √(\Delta))/(2a)=(-(-2) \pm 2√(3))/(2 * 2)=(2 \pm 2√(3))/(2 * 2)=(2(1 \pm √(3)))/(2 * 2)=(1 \pm √(3))/(2) \n\boxed{x=(1-√(3))/(2) \hbox{ or } x=(1+√(3))/(2)}$

7)
$x^2-x+2=0 \n \na=1 \n b=-1 \n c=2 \n\Delta=b^2-4ac=(-1)^2-4 * 1 * 2=1-8=-7 \n \n\boxed{\hbox{the discriminant } \Delta=-7}$

8)
$3x^2-6x+1=0 \n \na=3 \n b=-6 \n c=1 \n \Delta=b^2-4ac=(-6)^2-4 * 3 * 1=36-12=24 \n \n\boxed{\hbox{the discriminant } \Delta=24} \n \n\hbox{if } \Delta\ \textless \ 0 \hbox{ then there are no real roots} \n\hbox{if } \Delta=0 \hbox{ then there's one real root} \n\hbox{if } \Delta\ \textgreater \ 0 \hbox{ then there are two real roots} \n \n\Delta=24\ \textgreater \ 0 \n\boxed{\hbox{the equation has two real roots}}$

9)
$y=2x^2+x-3 \n \n a=2 \n b=1 \n c=-3 \n \Delta=b^2-4ac=1^2-4 * 2 * (-3)=1+24=25 \n \n \hbox{the function has two zeros} \n \boxed{\hbox{the parabola has 2 points in common with the x-axis}} \n \n a\ \textgreater \ 0 \hbox{ so the parabola ope} \hbox{ns upwards} \n \boxed{\hbox{the vertex lies below the x-axis}}$

10)
$y=x^2-12x+12 \n \na=1 \n b=-12 \n c=12 \n \Delta=b^2-4ac=(-12)^2-4 * 1 * 12=144-48=96 \n \n \hbox{the function has two zeros} \n \boxed{\hbox{the parabola has 2 points in common with the x-axis}} \n \n a\ \textgreater \ 0 \hbox{ so the parabola ope} \hbox{ns upwards} \n \boxed{\hbox{the vertex lies below the x-axis}}$

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Answers

Answer:

all of em

Step-by-step explanation:

Idk this sorry I can’t help

Rounak has 4 packs of 12 glitter pens. She has 7 packs of 8 scented pens. She gives her friend a pack of glitter pens in exchange for a new pack of scented pens. How many pens does Rounak have now?