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VMariaS [17]
3 years ago
10

Solve the following quadratic equation.

Mathematics
1 answer:
Viktor [21]3 years ago
3 0

Answer:

x = -7 and 7

Step-by-step explanation:

4x^2-10=186 add 10 on each side

4x^2 =196 divide by 4  on each side

x^2 = 49 take a √  on each side

x = -7,7

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The table shows the linear relationship between the
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Step-by-step explanation:

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3 years ago
An airplane flies due north from ikeja airport for 500km.It then flies on a bearing of 060 from a further distance of 300km befo
Nata [24]

Answer:

482 km

63.94 degrees

Step-by-step explanation:

to solve this question we will use the cosine rule. For starters, draw your diagram. From point A, up north is 500km and 060 from there, another 300. If you join the point from the road junction back to the starting point, yoou have a triangle.

Cosine rule states that

C = \sqrt{A^{2} + B^{2} -2AB   cos(c)  }

where both A and B are the given distances, 500 and 300 respectively, C is the 3rd distance we're looking for and c is the given angle, 060

solving now, we have

C = \sqrt{500^{2} + 300^{2} -2 * 500 * 300 cos(60)  }

C = \sqrt{250000 + 90000 - [215000   cos(60)  }]

C = \sqrt{340000 - [215000 * 0.5  }]

C = \sqrt{340000 - [107500  }]

C =\sqrt{232500}

C = 482 km

The bearing can be gotten by using the Sine Rule.

\frac{sina}{A} = \frac{sinc}{C}

sina/500 = sin60/482

482 sina = 500 sin60

sina = \frac{500 sin60}{482}

sina = 0.8983

a = sin^-1(0.8983)

a = 63.94 degrees

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3 years ago
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3 years ago
Make an equation with 123559 that equals 68.
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3 years ago
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