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3 years ago

For (triangle)DEF, you are required to construct the median DG to EF. Which of the following must be performed first?

Mathematics

1 answer:

iren2701 [21]3 years ago

8 0

Answer: C. Bisect DE at G.

Step-by-step explanation:

In a triangle, a median is a line segment drawn from a vertex to the midpoint of the opposite side of the vertex.

So, for triangle DEF , if we draw the median DG to the opposite side EF.

Here EF is the opposite side to vertex D.

So, G must be the midpoint of line segment EF.

So , we need to first mark G on EF.

i.e. we need to bisect EF at G.

So first step would be : Bisect EF at G.

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On a graph which line is horizontal and vertical? y or x

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Answer:

The Y is the vertical line and X is the horizontal line on the graph. So if you have point so (5,4). Five would be on the X-axis and 4 on Y-axis.

Step-by-step explanation:

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3 years ago

Reika [66]

The value of x will be equal to x = 79.

<h3>What is the triangle?</h3>

Triangle is a shape made of three sides in a two-dimensional plane. the sum of the three angles is 180 degrees.

For an isosceles triangle, the value of two opposite angles is equal and the opposite sides are also equal.

The Sum of the three angles is equal to 180 degrees so the third angle will be calculated as:-

Angle = 180 - 142 - 19 = 19

Now applying Lami's theorem in the triangle we will get the third side.

$\dfrac{sin\angle A}{a} = \dfrac{sin\angle B}{b} = \dfrac{sin\angle C}{c}$

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$x = \dfrac{42\times Sin 142}{Sin19}$

x = 79

Therefore the value of x will be equal to x = 79.

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2 years ago

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3 years ago

Circle P is described by the equation (x−1)2+(y+6)2=9 and circle Q is described by the equation (x+4)2+(y+14)2=4. Select from th

xeze [42]

Answer:

(a) Circle Q is 9.4 units to the center of circle P

(b) Circle Q has a smaller radius

Step-by-step explanation:

Given

$P:(x - 1)^2 + (y + 6)^2 = 9$

$Q:(x + 4)^2 + (y + 14)^2 = 4$

Solving (a): The distance between both

The equation of a circle is:

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

Where

$Center: (h,k)$

$Radius:r$

P and Q can be rewritten as:

$P:(x - 1)^2 + (y + 6)^2 = 3^2$

$Q:(x + 4)^2 + (y + 14)^2 = 2^2$

So, for P:

$Center = (1,-6)$

$r = 3$

For Q:

$Center = (-4,-14)$

$r = 2$

The distance between them is:

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

Where:

$Center = (1,-6)$ --- $(x_1,y_1)$

$Center = (-4,-14)$ --- $(x_2,y_2)$

So:

$d = \sqrt{(1 - -4)^2 + (-6 - -14)^2$

$d = \sqrt{(5)^2 + (8)^2$

$d = \sqrt{25 + 64$

$d = \sqrt{89$

$d = 9.4$

Solving (b): The radius;

In (a), we have:

$r = 3$ --- circle P

$r = 2$ --- circle Q

By comparison

$2 < 3$

Hence, circle Q has a smaller radius

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3 years ago