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3 years ago

8

A plane intersects a prism to form a cross section that is a polygon with five sides. The minimum number of sides that the polyg

on at the base of
the prism must have ls ..

1 answer:

antoniya [11.8K]3 years ago

4 0

Answer:

Since the plane if cutting through the prism to make 5 sides, than the original shape needs to have less than 5.

A prism, is a 3D shape, so you can't have a 2 sided shape.

Now you have a choice of 3+or 4 sides.

A 3 sides shape is a triangle, which also can't be cut into a 5 sided prism, so the answer would need to be 4.

Step-by-step explanation:

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1.<br> (3x + y = 5<br> (x-4y=32

chubhunter [2.5K]

Answer:

13x=52

x=14

y=-37

plz mark brainliest

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3 years ago

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Help me please, Find a, b, and c.

Mama L [17]

Answer:

a = $6*\sqrt{3}$

b = 12

c = $6\sqrt{2}$

Step-by-step explanation:

Since the triangles are right triangles with 60 and 45 degree angles, their side lengths follow special triangles.

A 45-45-90 right triangle has side lengths $1-1-\sqrt{2}$ .

A 30-60-90 right triangle has side lengths $1 - \sqrt{3} -2$ .

Starting with the top triangle which has a 60 degree angle, its side length 6 corresponds to a side length of 1 in the special triangle. It is 6 times bigger so its remaining sides will be 6 times bigger too.

Side a corresponds to side length $\sqrt{3}$ . Therefore, $a = 6*\sqrt{3}$ .

Side b corresponds to side length 2, b = 2*6 = 12.

The bottom triangle has a 45 degree angle, its side length b= 12 corresponds to $\sqrt{2}$ . This means $\sqrt{2}$ was multiplied by $12 = \sqrt{2} * \sqrt{72}$ . This means that side c is $\sqrt{72}=6\sqrt{2}$ .

3 0

3 years ago

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CosA×cos2A+sinA×sin2A=cos (2A-A)

Aleksandr-060686 [28]

Answer:

csc

A

, as desired!

Step-by-step explanation:

cos

2

A

sin

A

+

sin

2

A

cos

A

,

=

cos

2

A

cos

A

+

sin

2

A

sin

A

sin

A

cos

A

,

=

cos

(

2

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−

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sin

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cos

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cos

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sin

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cos

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7 0

3 years ago

Quadratics need help making parabola

timama [110]

Answer:

The equation of the parabola is $y = \frac{1}{3}\cdot x^{2}-\frac{4}{3}\cdot x -4$ , whose real vertex is $(x,y) = (2, -5.333)$ , not $(x,y) = (2, -5)$ .

Step-by-step explanation:

A parabola is a second order polynomial. By Fundamental Theorem of Algebra we know that a second order polynomial can be formed when three distinct points are known. From statement we have the following information:

$(x_{1}, y_{1}) = (-2, 0)$ , $(x_{2}, y_{2}) = (6, 0)$ , $(x_{3}, y_{3}) = (0, -4)$

From definition of second order polynomial and the three points described above, we have the following system of linear equations:

$4\cdot a -2\cdot b + c = 0$ (1)

$36\cdot a + 6\cdot b + c = 0$ (2)

$c = -4$ (3)

The solution of this system is: $a = \frac{1}{3}$ , $b = - \frac{4}{3}$ , $c = -4$ . Hence, the equation of the parabola is $y = \frac{1}{3}\cdot x^{2}-\frac{4}{3}\cdot x -4$ . Lastly, we must check if $(x,y) = (2, -5)$ belongs to the function. If we know that $x = 2$ , then the value of $y$ is:

$y = \frac{1}{3}\cdot (2)^{2}-\frac{4}{3}\cdot (2) - 4$

$y = -5.333$

$(x,y) = (2, -5)$ does not belong to the function, the real point is $(x,y) = (2, -5.333)$ .

5 0

3 years ago

Factor -1/3 out of -1/3x - 12

zaharov [31]

Answer:

-1/3(x+36)

Step-by-step explanation:

4 0

3 years ago