Please help with number 10

Mathematics

1 answer:

Snezhnost [94]3 years ago

7 0

Answer:

Step-by-step explanation:

XY = 6B

YZ= 8B

XZ = 175

XYZ= 6B + 8B

XYZ = 14B

XYZ = XZ = 175 = 14B

14B=175

14B/14 =175/14

B=12.5

XY = 6B = 6 * 12.5 = 75

YZ= 8B = 8 * 12.5 = 100

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Where would an imaginary line need to be drawn to reflect across an axis of symmetry so that a regular pentagon can carry onto i

oksano4ka [1.4K]

The correct answer is:

A line of reflection would need to be drawn as a line that bisects a vertex angle of the pentagon.

Explanation:

A line that bisects a vertex angle of a pentagon will be a line of symmetry within the pentagon. The pentagon can be "folded" through this line into two equal halves.

This imaginary line represents the center of the pentagon. A reflection through this line will carry the pentagon onto itself. This is due to the definition of line symmetry:

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3 years ago

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In the standard x,y plane, a circle has a radius 6 and center (7, 3). The circle intersects the x-axis at (a, 0) and (b, 0). Wha

arsen [322]

Answer: The required value of a+b is 14.

Step-by-step explanation: Given that in the standard xy plane, a circle has a radius 6 and center (7, 3).

The circle intersects the x-axis at (a, 0) and (b, 0).

We are to find the value of a+b.

We know that

the standard equation of a circle with center at (h, k) and radius r units is given by

$(x-h)^2+(y-k)^2=r^2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

For the given circle, we have

(h, k) = (7, 3) and r = 6 units.

So, from equation (i), we get

$(x-7)^2+(y-3)^2=6^2\\\\\Rightarrow (x-7)^2+(y-3)^2=36~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Since the circle (ii) passes through the points (a, 0) and (b, 0), so let the point be denoted by (c, 0), then we have

$(c-7)^2+(0-3)^2=36\\\\\Rightarrow (c-7)^2+9=36\\\\\Rightarrow (c-7)^2=27\\\\\Rightarrow c-7=\pm3\sqrt3~~~~~~~~~~~~~~~~~~~~~~[\textup{Taking square root on both sides}]\\\\\Rightarrow c=7\pm3\sqrt3$