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Oduvanchick [21]

2 years ago

8

Someone plz help I rlly need to finish

1 answer:

bekas [8.4K]2 years ago

6 0

Since TSQ and QSR are supplementary, they give 180 when summed. TSQ is 150, so QSR must be 30.

Therefore, you have

$\tan(30)=\dfrac{\sin(30)}{\cos(30)}=\dfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \dfrac{1}{2}\cdot\dfrac{2}{\sqrt{3}}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}$

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A line that passes through 2 or more lines are called a?<br>

11Alexandr11 [23.1K]

Answer:

transversal

Step-by-step explanation:

1. In geometry any line which passes through or intersects 2 or more line are called a transversal.

2.Transversal are generally used in geometry of Euclidean plane to decide whether the given set of lines through which transversal passes are parallel or not.

8 0

2 years ago

Which can be a next step in the construction of an angle with a side on line l that is congruent to ∠ABC?

iren2701 [21]

A) use a straightedge to draw a line thought D and F

6 0

2 years ago

A rectangular box has a square base. The combined length of a side of the square base, and the height is 20 in. Let x be the len

aniked [119]

Answer:

a. V = (20-x) $x^{2}$ $in^{3}$

b . 1185.185 $in^{3}$

Step-by-step explanation:

Given that:

The height: 20 - x (in )
Let x be the length of a side of the base of the box (x>0)

a. Write a polynomial function in factored form modeling the volume V of the box.

As we know that, this is a rectangular box has a square base so the Volume of it is:

V = h * $x^{2}$ $in^{3}$

<=> V = (20-x) $x^{2}$ $in^{3}$

b. What is the maximum possible volume of the box?

To maximum the volume of it, we need to use first derivative of the volume.

<=> dV / Dx = -3 $x^{2}$ + 40x

Let dV / Dx = 0, we have:

-3 $x^{2}$ + 40x = 0

<=> x = 40/3

=>the height h = 20/3

So the maximum possible volume of the box is:

V = 20/3 * 40/3 *40/3

= 1185.185 $in^{3}$

7 0

2 years ago

Two cats, Fluffy and Fireball, met in the park. Fluffy's tail is 1/3, of a meter long. Fireball's tail is 1/4 of a meter long.

Lunna [17]

1/3 - 1/4 = 4/12 - 3/12 = 1/12....Fluffy's tail is 1/12 meters longer then Fireball's tail

6 0

2 years ago

Read 2 more answers

Line segment NY has endpoints N(-11, 5) and Y(3,-3).

777dan777 [17]

Given:

Line segment NY has endpoints N(-11, 5) and Y(3,-3).

To find:

The equation of the perpendicular bisector of NY.

Solution:

Midpoint point of NY is

$Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

$Midpoint=\left(\dfrac{-11+3}{2},\dfrac{5-3}{2}\right)$

$Midpoint=\left(\dfrac{-8}{2},\dfrac{2}{2}\right)$

$Midpoint=\left(-4,1\right)$

Slope of lines NY is

$m=\dfrac{y_2-y_1}{x_2-x_1}$

$m=\dfrac{-3-5}{3-(-11)}$

$m=\dfrac{-8}{14}$

$m=\dfrac{-4}{7}$

Product of slopes of two perpendicular lines is -1. So,

$m_1\times \dfrac{-4}{7}=-1$

$m_1=\dfrac{7}{4}$

The perpendicular bisector of NY passes through (-4,1) with slope $\dfrac{7}{4}$ . So, the equation of perpendicular bisector of NY is

$y-y_1=m_1(x-x_1)$

$y-1=\dfrac{7}{4}(x-(-4))$

$y-1=\dfrac{7}{4}(x+4)$

$y-1=\dfrac{7}{4}x+7$

Add 1 on both sides.

$y=\dfrac{7}{4}x+8$

Therefore, the equation of perpendicular bisector of NY is $y=\dfrac{7}{4}x+8$ .

6 0

2 years ago