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natta225 [31]
3 years ago
12

Find the m slope in 8x+10y=11

Mathematics
1 answer:
skelet666 [1.2K]3 years ago
8 0

Answer: The answer is -4/5 hope this helps :D

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Simplify: a+2a+3a+4a
kolezko [41]

Since all of these numbers have the same variable they can all be added up to get a sum of 10a which is its simplified form.

5 0
2 years ago
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Find the height of a square pyramid that has a volume of 12 cubic feet and a base length of 3 feet
Andreyy89
I got the height to be 4
8 0
3 years ago
Your school's drama club charges $4 per student for admission to Our Town. The club borrowed $400 from parents to pay for costum
inna [77]

Answer:

125 students attend.

Step-by-step explanation:

We already know that the drama club made $500 (500 - 400 = 100)

divide 500 by 4 to find out how many students attend the play, and paid $4

500 ÷ 4 = 125

125 students attend.

6 0
3 years ago
Use the quadratic formula to solve the equation. -2x^2-x+7=0.
Furkat [3]
<span>-2x^2-x+7=0  
Variable with the highest degree's (exponent) constant, -2 is a, next variable's constant, -1 is b, the constant or number without a variable, 7 is c

using substitution put the numbers into the formula
</span>(-b±√(b^(2)-4ac))/(a^(2))

(-(-1)±√((-1)^(2)-4(-2)(7))/((-2)^(2)) simplify 

(1±√(1+56))/4

1±√(57)/4 is your answer
8 0
3 years ago
X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?
cluponka [151]

Answer:

x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

Step-by-step explanation:

I'm assuming the system is \left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.:

x^2+y^2=2

x^2+(2x^2-3)^2=2

x^2+(4x^4-12x^2+9)=2

x^2+4x^4-12x^2+9=2

4x^4-11x^2+9=2

4x^4-11x^2+7=0

x^4-11x^2+28=0

(x^2-7)(x^2-4)=0

(4x^2-7)(x^2-1)=0

4x^2-7=0

4x^2=7

x^2=\frac{7}{4}

x=\pm\sqrt{\frac{7}{4}}

x^2-1=0

x^2=1

x=\pm1

y=2x^2-3

y=2(\pm\sqrt{\frac{7}{4}})^2-3

y=2({\frac{7}{4}})-3

y=\frac{7}{2}-3

y=\frac{1}{2}

y=2x^2-3

y=2(\pm1)^2-3

y=2(1)-3

y=2-3

y=-1

Therefore, x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

4 0
2 years ago
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