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saw5 [17]
2 years ago
11

1/3a = −5 What does a equal?

Mathematics
1 answer:
Natali5045456 [20]2 years ago
3 0
1/3 ⋅ = −5

1. Combine multiplied terms into a single fraction
1/3 = −5
1a/3 = -5

2. Multiply by 1
1a/3 = -5
a/3 = -5

3. Multiply all terms by the same value to eliminate fraction denominators
a/3 = -5
3 • a/3 = 3 (-5)

4. Cancel multiplied terms that are in the denominator
3 • a/3 = 3 (-5)
a = 3 (-5)

5. Multiply the numbers
a = 3(-5)
a = -5
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put back together
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Step-by-step explanation:

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2 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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