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3 years ago

11

Tom gets $12 off a box of chocolates that had an original priceof $48. What percentage is the discount?

2 answers:

Crank3 years ago

4 0

His discount is 12% off the original price.

BabaBlast [244]3 years ago

4 0

The discount percentage is %25

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Can someone help me please

alekssr [168]

Use the formula x+x+1+x+2 or 3x+3=170 to find x. x is the first number. Then add x+1 to find the second number. Then add x+2 to find the third final number.
x will equal 167/3 which is a decimal, so just plug it in to the 3 equations I mentioned above to find your 3 consecutive numbers. More questions? Just ask me!

4 0

3 years ago

Read 2 more answers

Cari is searching online for airline tickets. Two weeks ago, the cost to fly from Boston to Denver was $225. Now the cost is $30

liq [111]

Answer:

33.33% increase. 55.56% increase.

Step-by-step explanation:

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2 years ago

Tobeg says the equation 3x + 2y= 270 and y=4/5x + 20 intersect at the point (50,60) is this true how do you know

il63 [147K]

ANSWER

Yes it is very true

<u>EXPLANATION</u>

If the two equations intersect at $(50,60)$ then this point must satisfy the two equations.

$3x+2y=270---(1)$

We substitute $(50,60)$ in to erquation (1)

$3(50)+2(60)=270$

$150+120=270$

$270=270$

$y=\frac{4}{5}x+20 ---(2)$

We now substitute $(50,60)$ in to erquation (2) also

$60=\frac{4}{5}(50)+20$

$60=40+20$

$60=60$

Since the point satisfy all the two equations, it is true that they intersect at $(60,50)$

3 0

3 years ago

Find an equation of a line that comes close to the points listed in the table.

sergey [27]

$y=x+2\\x=6\to y=6+2=8\\x=10\to y=10+2=12\\x=12\to y=12+2=14\\x=16\to y=16+2=18\\x=18\to y=18+2=20\\-----------------------\\y=x+3\\x=6\to y=6+3=9\\x=10\to y=10+3=13\\x=12\to y=12+3=15\\x=16\to y=16+3=19\\x=18\to y=18+3=21\\-----------------------\\Answer:y=x+1.5$

3 0

2 years ago

Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

7 0

2 years ago