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Harrizon [31]
3 years ago
10

How do you simply the square root of 28

Mathematics
1 answer:
maria [59]3 years ago
3 0
\sqrt{28} can be rewritten as ;
=\sqrt{7*4}
=\sqrt{4} * \sqrt{7}
=2\sqrt{7}
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Robyn had 8 apples. She ate3/8 of them. What is the number of apples she ate?
Leni [432]
Since she only has 8 apples and she ate 3 of 8, she ate 3 apples.
8 0
3 years ago
Read 2 more answers
Lisa has $7.80 to spend on some tomatoes and a loaf of bread. Tomatoes
Sunny_sXe [5.5K]

Answer:

Lisa can buy 5 pounds of tomatoes.

Step-by-step explanation:

Given that Lisa has $ 7.80 to spend on some tomatoes and a loaf of bread, and tomatoes cost $ 1.20 per pound, and a loaf of bread costs $ 1.80, to determine how many pounds of tomatoes she can buy, the following calculation must be performed:

1.20X + 1.80 = 7.80

1.20X = 7.80 - 1.80

X = 6 / 1.20

X = 5

So Lisa can buy 5 pounds of tomatoes.

8 0
3 years ago
The made 9 pounds of fudge . The fudge was separated into 3/4 pound blocks . They sell each block for $6.50 .if they sell all th
Flauer [41]
Number 16, he put 1/2 of his money into his savings and took the rest to spend. So fraction of his amusement park money he spent on rides and popcorn is (1/5 + 3/4) x 1/2 = 19/40 <=> A

Each 3/4 pound block for $6.50. So each pound block will be $6.50 x 4/3 = 26/3. They have 9 pounds of fudge, money they will make is 26/3 x 9 = $78
3 0
3 years ago
2x^2+5x+3/x^2-3x-4 divided by 4x^2+2x-6/x^2-8x+16
Masja [62]

Answer: \frac{x-4}{2x-2}

Step-by-step explanation:

\frac{2x^2+5x+3}{x^2-3x-4} divided by \frac{4x^2+2x-6}{x^2-8x+16} is the same thing as multiplying \frac{2x^2+5x+3}{x^2-3x-4} by \frac{x^2-8x+16}{4x^2+2x-6}.

The equation we get is:

\frac{2x^2+5x+3}{x^2-3x-4}*\frac{x^2-8x+16}{4x^2+2x-6}

With this equation, we can factor each equation:

\frac{(x+1)(2x+3)}{(x+1)(x-4)} *\frac{(x-4)(x-4)}{2(x-1)(2x+3)}

We can cancel out like terms since they would be dividing each other:

\frac{x-4}{2x-2}

8 0
3 years ago
Please solve this sheet.
lyudmila [28]
Hello,

A: roots: -1,-3
a point (-2,1)
Vertex=((-2,1)

y=k*(x+1)(x+3) using roots
but k*(-2+1)(-2+3)=1==>k*(-1)*1=1==>k=-1

eq: y=-(x+1)(x+3)
==>y=-(x²+3x+x+3)
==>y=-x²-4x-3
y=k(x+2)²+1 if x=-1,y=0 ==>k*1+1=0==>k=-1
==>y=-(x+2)²+1

Answer :A--> R,K

B)
y=k(x+4)²-2 and k=-1/2
y=-1/2(x+4)²-2
y=-1/2x²-4x-10

answer B--> I,≈W if it is written -1/2*x²  (square has been forgotten)

C:
y=2x²-16x+30
y=2(x-4)²-2
answer : C-->S,J

D:
y=-(x+3)(x+1)
y=-x²-4x-3
=-(x+2)²+1
answer D--> V,L

E:
Here there is a problem: or the graph is wrong, or 2 equations are missing!

y=1(x+1)(x-3) using roots
y=x²-2x-3 ≈ T si it were -2x and not +2x.

y=(x-1)²-4 ≈H is it were -1 in place of +1 [H:y=(x+1)²-4]




6 0
3 years ago
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