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3 years ago

10

4

1 answer:

Orlov [11]3 years ago

5 0

Answer:

B. on edge2021

Step-by-step explanation:

Just did it

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The area of the triangle is<br><br> square inches.<br> fill in the blanks

Dvinal [7]

Answer:

48sq^2

Step-by-step explanation:

(12*8)/2

96/2

48

3 0

3 years ago

HELP PLEASE!! WILL GIVE BRAINLIEST!!!

beks73 [17]

Answer:

Slope = 2 Y-Intercept = (0, 3)

Step-by-step explanation:

Slope : $\frac{Rise}{Run}$ which is $\frac{y2 - y1}{x2 - x1}$ from two points: point1 (x1 , y1) and point2 (x2, y2).

For the purpose of getting the slope I will be using point (0, 3) as point 1

and point (-3, -3) as point 2.

If you plug in 0 as x1, 3 as y2, -3 for x2, and -3 for y2:

$\frac{-3 - 3}{-3 - 0}$ = $\frac{-6}{-3}$ = 2

Y-Intercept: The y intercept is the point where the line is touching the y axis. Which in this graph, there is only point (0,3).

7 0

3 years ago

trasher [3.6K]

Answer:

See explanation

Step-by-step explanation:

The given complex number are:

$z_1=-3\sqrt{3}+3i$

and

$z_2=6\cos 150\degree+6i\sin 150\degree$

When we rewrite $z_1=-3\sqrt{3}+3i$ in complex form, we obtain;

$z_1=r(\cos \theta+i\sin \theta)$

where

$r=\sqrt{(-3\sqrt{3})^2+3^2 }=\sqrt{36}=6$

and

$\theta=tan^{-1}(\frac{y}{x})$

$\implies \theta=tan^{-1}(\frac{-3\sqrt{3}}{3})=150\degree$

Hence,

$z_1=6(\cos 150\degree+i\sin 150\degree)$

$z_1=6\cos 150\degree+6i\sin 150\degree$

Hence

$z_1=z_2$

8 0

3 years ago

Paying 20$ cashapp if you get this correct

Ilia_Sergeevich [38]

Answer:

148⁰

Step-by-step explanation:

M< EDF = M< DFE (because the two angles are equal because the triangle has two equal sides)

2x + 8 = 3x + 4

8 – 4 = 3x - 2x

4=x or x=4

So M< EDF= 2x4+8 = 16⁰

So M< EDF + M< DFE= 32⁰

180⁰- 32⁰ = 148⁰

3 0

3 years ago

NEED HELP ASAP !! WILL MARK BRAINLEAST<br> SHOW WORK TOO !!!

irinina [24]

Step-by-step explanation:

$(1 + \cos(x) )(1 - \cos(x) )$

$1 - \cos {}^{2} (x)$

$= \sin {}^{2} (x)$

b.

$\frac{1}{ \cot {}^{2} (x) } - \frac{1}{ \cos {}^{2} (x) }$

$\frac{1}{ \frac{ \cos {}^{2} (x) }{ \sin { }^{2} (x) } } - \frac{1}{ \cos {}^{2} (x) }$

$\frac{1}{ \cot {}^{2} (x) } - \frac{ \csc {}^{2} (x) {} }{ \cot {}^{2} (x) }$

$\frac{ - \cot {}^{2} (x) }{ \cot {}^{2} (x) }$

$- 1$

c.

$\sec {}^{2} ( \frac{\pi}{2} - x ) )( \sin {}^{2} (x) - \sin {}^{4} (x))$

$( \csc {}^{2} (x) )( \sin {}^{2} (x) - \sin {}^{4} (x )$

$\csc {}^{2} (x) ( \frac{1}{ \csc {}^{2} (x) } - \frac{1}{ \csc {}^{4} (x) } )$

$1 - \frac{1}{ \csc {}^{2} (x) }$

$1 - \sin {}^{2} (x)$

$\cos {}^{2} (x)$

3 0

2 years ago