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svp [43]
2 years ago
10

A recipe for lasagna calls for cups of cheese for the filling, plus cup of cheese for the top. The expression can be used to fin

d how much cheese is
needed to make 3 lasagnas. How many cups of cheese are needed for 3 lasagnas?
Mathematics
2 answers:
Mashutka [201]2 years ago
8 0
I don’t know sorrryyy
Svetradugi [14.3K]2 years ago
4 0

Answer:

do they have anymore to the question. is there a second part of the question???

Step-by-step explanation:

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I need help fast thanks
Dmitry [639]
-18x-12=4
Iam not sure if that’s what you want but basically I multiplied the numbers inside the bracket by -6
4 0
3 years ago
A school buys games for 6 classrooms. It buys 3 board games 4 puzzles games and 1 video game for each class room how many games
Snowcat [4.5K]

3+4+1 for each class so it's 8

8 x 6 for all the classrooms so it's 48.  The school buys 48 games

6 0
3 years ago
Read 2 more answers
What is the value of this expression?<br><br> ​ 1.5(−2.4+(−5.3))
Ahat [919]

Answer:

-11.55

Step-by-step explanation:

(to get the numbers out of the brackets you have to multiply all the numbers in the brackets to the number that is there, and in this case its 1.5)

1.5x(-2.4)= -3.6

1.5×(-5.3)= -7.95

after that, you add them (because there is a plus sign, not a minus, multiplication sign or a division sign.)

-7.95+ -3.6= -11.55

answer= -11.55

7 0
3 years ago
What is the slope of (9,6) (21,14)
Delvig [45]
You would write m= y2 -y1/x2-x1 so the answer would be 8/12
5 0
3 years ago
Read 2 more answers
Solve the inequality 2x&gt;30+5/4x
insens350 [35]

Answer:

Step-by-step explanation:

2x > 30+\frac{5}{4x} \\2x-\frac{5}{4x} > 30\\\frac{8x^2-5}{4x} > 30\\case~1\\if~x > 0\\8x^2-5 > 120x\\8x^2-120x > 5\\x^2-15x > \frac{5}{8} \\adding~(-\frac{15}{2} )^2~to~both~sides\\(x-\frac{15}{2} )^2 > \frac{5}{8}+\frac{225}{4} \\(x-\frac{15}{2} )^2 > \frac{455}{8} \\x-\frac{15}{2} < -\sqrt{\frac{455}{8} }  \\x < \frac{15}{2}-\sqrt{\frac{455}{8} } \\or~x < 0\\rejected~as~x > 0

x-\frac{15}{2} > \sqrt{\frac{455}{8} } \\x > \frac{15}{2} +\sqrt{\frac{455}{8} }

case~2

if~x < 0\\8x^2-5 < 120x\\8x^2-120x < 5\\x^2-15x < \frac{5}{8} \\adding~(-\frac{15}{2} )^2\\(x-\frac{15}{2} )^2 < \frac{5}{8} +(-\frac{15}{2} )^2\\|x-\frac{15}{2} | < \frac{5+450}{8} \\-\sqrt{\frac{455}{8} } < x-\frac{15}{2} < \sqrt{\frac{455}{8} } \\\frac{15}{2} -\sqrt{\frac{455}{8} } < x < \frac{15}{2} +\sqrt{\frac{455}{8} } \\but~x < 0\\7.5-\sqrt{\frac{455}{8} } < x < 0

8 0
2 years ago
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