Burger King calories 320Del Taco calories 280
Mcdonalds calories 280
Jack in the box calories 250
Wendy's calories 270

Find the mean number of calories for the hamburgers.
and the median number of calories for the hamburgers

Answers

Answer 1

Answer: arrange them from least to greatest form

250 cal., 270 cal., 280 cal., 280 cal., 320 cal.,

Mean = $(250+270+280+280+320)/(5)$
(5 because there is a total of 5 fast food restaurants' hanburgers)

= $(1400)/(5)$
Mean = 280 calories

Median = The number exactly in between,
In midst of these numbers, 280 is in the mid-way
Median = 280 calories

3/4

Hope this helps you.

For every penny Sam puts into his bank, Tara puts 4 pennies into her bank. If Sam puts 10 pennies into his bank, how many pennies does Tara put into her bank? Answer options with 4 options

Answers

Answer: If Tara puts 4 pennies in for every one Sam does then after he puts in 10 she would of put 40.

She put in 40 pennies. Hope this helps ;)

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

Solve the ecuation: $x=\frac{[x]+2012}{\{x\}+2013}$

where [x] is the whole part of the number x and {x} is the fractional part of the number x.

Answers

Scrii x = [ x ] + { x } ;
Folosesti proprietatea fundamentala a proportiilor;
Apoi, exprimi [ x ] in functie de { x }, si obtii ca :
[ x ] = ( 2012 - { x } ^2 - 2013{ x } ) / ( 2012 + { x } ) ;
Fie { x } = t ∈ [ 0 , 1 ) ;
Obtii ca [ x ] = - t - 1 + 4024 / ( t + 2012 ) ;
Stim ca [ x ] ∈ Z ;
Atunci t = 0 sau t = 2012 ;
Convine numai t = 0 => x = [ x ] ;
Ecuatia originala va avea solutia 1.

Bafta!

Complete this sentence: After a congruence transformation, the perimeter of a triangle would be _________________ it was before.A. cannot be determined
B. the same as
C. less than
D. greater than

Answers

Congruence transformation results to a perimeter of any polygon that is the same as the original dimensions and shape. In this case, this is similar to translation where all dimensions, angles are retained constant. The answer to this problem hence is B

Answer:the same as

Step-by-step explanation:

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