In expanded form, in kilometers, how far is Jupiter from the Sun?

(20pts=5pts each) For parts (a)–(d) answer the questions and complete the statements in the space provided: line EC ∩ line AC ∩

Lyrx [107]

Answer:

Step-by-step explanation:

Since you didn't state what you're looking for, I will lost out all the angles in the picture instead.

Given that

∠CEF = 150°

∠ECA = 32°

∠CEA = 180 - ∠CEF(angles in a straight line)

∠CEA = 180 - 150

∠CEA = 30°

∠EAC = 180 - ∠CEA - ∠ECA(angles in a triangle)

∠EAC = 180 - 30 - 32

∠EAC = 118°

∠CAB = 180 - ∠EAC(angles in a triangle)

∠CAB = 180 - 118

∠CAB = 62°

∠ACB = ∠CEA = 30°(similar angles)

∠ABC = 180 - ∠CAB - ∠ACB(angles in a triangle)

∠ABC = 180 - 62 - 30

∠ABC = 88°

∠ECB = ∠ECA + ∠ACB(angles in a triangle)

∠ECB = 32 + 30

∠ECB = 62°

∠CEA + ∠ECB + ∠ABC = 180

30 + 62 + 88 = 180°.

Please give brainliest if it helped you <3

5 0

2 years ago

If x+y=12 and xy=15,find the value of (x^2+y^2)

Mice21 [21]

Answer: $x^2+y^2=\dfrac{105\pm 18\sqrt6}{2}$

<u>Step-by-step explanation:</u>

EQ1: x + y = 12 --> x = 12 - y

EQ2: xy = 15

Substitute x = 12-y into EQ2 to solve for y:

(12 - y)y = 15

12y - y² = 15

0 = y² - 12y + 15

↓ ↓ ↓

a=1 b= -12 c=15

$.\ y=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\.\quad =\dfrac{-(-12)\pm \sqrt{(-12)^2-4(1)(15)}}{2(1)}\\\\\\.\quad =\dfrac{12\pm \sqrt{144-120}}{2}\\\\\\.\quad =\dfrac{12\pm \sqrt{24}}{2}\\\\\\.\quad =\dfrac{12\pm 2\sqrt{6}}{2}\\\\\\.\quad =6\pm \sqrt{6}$

Now, let's solve for x:

$xy=15\\\\x(6\pm\sqrt6)=15\\\\x=\dfrac{15}{6\pm\sqrt6}\\\\\\x=\dfrac{15}{6\pm\sqrt6}\bigg(\dfrac{6\pm\sqrt6}{6\pm\sqrt6}\bigg)=\dfrac{6\pm \sqrt6}{2}$

Lastly, find x² + y² :

$y^2=(6\pm \sqrt6)^2\quad \rightarrow \quad y^2=36\pm 12\sqrt6 +6\quad \rightarrow \quad y^2=42\pm 12\sqrt6$

$x^2=\bigg(\dfrac{6\pm \sqrt6}{2}\bigg)^2\quad \rightarrow \quad x^2=\dfrac{42\pm 12\sqrt6}{4}\quad \rightarrow \quad x^2=\dfrac{21\pm 6\sqrt6}{2}$

$x^2+y^2=\dfrac{21\pm 6\sqrt6}{2}+42\pm 12\sqrt6\\\\\\.\qquad \quad = \dfrac{21\pm 6\sqrt6}{2}+\dfrac{84\pm 24\sqrt6}{2}\\\\\\. \qquad \quad = \dfrac{105\pm 18\sqrt6}{2}$

5 0

3 years ago

What is a quadratic function?

ycow [4]

C - F(x)= 5x^2-4x+5

Quadratic function equations have an x raised to the second power and makes a parabola when graphed. B could potentially also be right depending on who you ask because x is being raised to the second power but since the equation says 0x it means there is no x to be raised to the second power so i would say C.

3 0

2 years ago

If tan A= 1,what is the value of A

xxMikexx [17]

Answer:

45°

Step-by-step explanation:

Ok, so, there is one thing I need to point out. 45° is the 'main' value if you assume 0°<A<180°. However, sin, cos, and tan have different periods which means that there are infinite values of A where tanA = 1. The general notation that you could put is A = 45° + (n*180°) where n is just a number. For example, if n = 1, you would get an angle of 225°. If you plug tan225° into the calculator, you get 1. If you did radians, you could write A = $\frac{\pi}{2} + n\pi$ . But ignore that if you haven't. Basically, the answer would be 45° if you are assuming A is between 0° and 180°. Also, you could have just used your calculator and types inverse tan function ( $tan^{-1}$ ) and plug in 1 to find the primary answer of 45.

6 0

2 years ago

It costs $5 per hour to rent a snowboard from a certain ski rental company, plus a $50 deposit. Another ski rental company charg

Leokris [45]

For the first part
5 an hour plus 50
Can you make an equation out of it

5 0

2 years ago

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