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3 years ago

Use the figure to determine the intersection of planes EAB and EFG

Mathematics

1 answer:

vredina [299]3 years ago

7 0

Answer:

Option (3). EF

Step-by-step explanation:

From the figure attached,

Plane defined by EAB can be represented by the face EABF of the square prism also.

Similarly, plane EFG can be represented by the face EFGH of the prism.

Now these sides EABF and EFGH are joining each other at the edge EF of the cuboid.

Therefore, intersection of the given planes is EF.

Option (3) will be the answer.

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Answer:0.26

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2 years ago

your sister is considering two different shapes for her garden. one is a square with side lengths of 3.5 meters, and the other i

In-s [12.5K]

The area of a square: (3.5m)²=12.25m²
The area of a circle: π*(2m)²=12.57m²
The circle shape of garden will give my sister more space to plant.

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2 years ago

A 14-foot ladder is placed against a vertical wall of a building, with the bottom of the ladder standing on level ground 9 fee

timama [110]

Answer:

10.7 feet

Step-by-step explanation:

The ladder, the ground and the wall form the shape of a right angled triangle as shown in the image below.

The hypotenuse of the triangle is 14 feet (length of ladder)

The base of the triangle is 9 feet long (the distance from the base of the ladder to the wall)

We need to find the height of the triangle. We can apply Pythagoras rule:

$hyp^2 = a^2 + b^2$

where hyp = hypotenuse

a = base of the triangle

b = height of the triangle

Therefore:

$14^2 = 9^2 + b^2\\\\196 = 81 + b^2\\\\b^2 = 196 - 81 = 115\\\\b = \sqrt{115} \\\\b = 10.7 feet$

The wall reaches 10.7 feet high.

3 0

3 years ago

Solve the equation: k^2+5k+13=0

mr_godi [17]

Step-by-step explanation:

k² + 5k + 13 = 0

Using the quadratic formula which is

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\$

From the question

a = 1 , b = 5 , c = 13

So we have

$k = \frac{ - 5 \pm \sqrt{ {5}^{2} - 4(1)(13) } }{2(1)} \\ = \frac{ - 5 \pm \sqrt{25 - 52} }{2} \\ = \frac{ - 5 \pm \sqrt{ - 27} }{2} \: \: \: \: \: \: \\ = \frac{ - 5 \pm3 \sqrt{3} \: i}{2} \: \: \: \: \: \:$

<u>Separate the solutions</u>

$k_1 = \frac{ - 5 + 3 \sqrt{3} \: i }{2} \: \: \: \: or \\ k_2 = \frac{ - 5 - 3 \sqrt{3} \: i}{2}$

The equation has complex roots

<u>Separate the real and imaginary parts</u>

We have the final answer as

$k_1 = - \frac{5}{2} + \frac{3 \sqrt{3} }{2} \: i \: \: \: \: or \\ k_2 = - \frac{5}{2} - \frac{3 \sqrt{3} }{2} \: i$

Hope this helps you

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2 years ago

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Amiraneli [1.4K]

Answer:

Step-by-step explanation:

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2 years ago