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Charra [1.4K]
3 years ago
15

At first,Ben had $90 and Chandra had $48 .each bought a shirt at the same price.the amounts of money Ben and chandra had left we

re in the ratio 4:1 .how much did the shirt cost?
Mathematics
1 answer:
antiseptic1488 [7]3 years ago
7 0
90-48 = 42
3 units = 42
1 unit = 14
48-14=34
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It costs Doug $30 to set up a cupcake stand. He charges $2 for every cupcake he sells. If Doug makes $48 of profit in one day, h
Lunna [17]

Answer:

It is 39 cupcakes.

Since he charges $2 per cupcake (39), it is 78.

He makes profit so subtract 30 from 78 and you get $48.

6 0
3 years ago
Free points: this is because of all the kindness I have seen on brainly:
const2013 [10]

Answer:

10x10= 100

Step-by-step explanation:

thx for the points

hope you have a wonderful dayyy

3 0
3 years ago
Read 2 more answers
The hypotenuse of a right triangle is 10cm long. The longer leg is 2cm longer than the shorter leg. Find the side lengths of the
Crazy boy [7]
We need Pythagoras theorem here
a^2+b^2 = c^2
a, b =  legs of a right-triangle
c = length of hypotenuse

Let S=shorter leg, in cm, then longer leg=S+2 cm
use Pythagoras theorem
S^2+(S+2)^2 = (10 cm)^2
expand (S+2)^2
S^2 +   S^2+4S+4  = 100 cm^2   [collect terms and isolate]
2S^2+4S =  100-4 = 96 cm^2
simplify and form standard form of quadratic
S^2+2S-48=0
Solve by factoring
(S+8)(S-6) = 0   means (S+8)=0, S=-8
or                                   (S-6)=0, S=6
Reject nengative root, so
Shorter leg = 6 cm
Longer leg = 6+2 cm = 8 cm
Hypotenuse (given) = 10 cm

5 0
3 years ago
Y = (x + 2)^2-3 in standard form?
Verdich [7]
Y=2x2−8x+5 is standard form
4 0
2 years ago
Which set of coefficients of the terms in the expansion of the binomial (x+y)^3 is correct ?
BigorU [14]
We are technically FOILing this out... with a power of 3.

(x+y)(x+y)(x+y)

So we can first factor out the first two "x+y"s.

( x^{2} +2xy + y^{2} ), multiplied by the last "x+y".

x^{3} +3 x^{2} y +3x y^{2} + y^{3}

Coefficients are the number that comes in front of a variable. 

In this case, 1 comes in front of x^{3}, 3 comes in front of x^{2} y, 3 comes in front of x y^{2}, and 1 comes in front of y^{3}.

Thus: 1, 3, 3, 1.
         Answer Choice A
4 0
3 years ago
Read 2 more answers
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