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ValentinkaMS [17]
3 years ago
8

Find the value of x. Part 1c​

Mathematics
2 answers:
aniked [119]3 years ago
6 0

Answer: x = 11

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

a) bottom triangle, bottom right angle: Using Linear Pair Postulate, 180° - 114° = 66°

b) bottom triangle, top angle: Congruent sides implies congruent angles = 66°

c) bottom triangle, bottom left angle: Using the Triangle Sum Theorem, 66° + 66° + y = 180° --> y = 48°

d) top triangle, bottom angle: Using Complimentary Angles Theorem, 90° - 48° = 42°

Congruent sides implies congruent angles:

42° = ∠2

42 = 2x + 20

22 = 2x

11 = x

gtnhenbr [62]3 years ago
4 0

Answer:

x=11

Step-by-step explanation:

Please see attached image.

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3x+10 in a word phrase
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When put into a word phrase 3x+10 is three of x plus ten, three x plus ten etc.
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3 years ago
The quotient of 20 and 3 more than x is 4. Find x.
tigry1 [53]

Answer:

x=2

Step-by-step explanation:

Quotient means division

20/(x+3) =4

Multiply each side by (x+3)

(x+3)*20/(x+3) =4(x+3)

20 = 4(x+3)

Divide each side by 4

20/4 = 4(x+3)/4

5 = x+3

Subtract 3 from each side

5-3 = x+3-3

2 =x

4 0
2 years ago
Read 2 more answers
Solve using the quadratic formula.
Ivanshal [37]

the answer is z=0, 1 over 3

7 0
3 years ago
A thin metal plate, located in the xy-plane, has temperature T(x, y) at the point (x, y). Sketch some level curves (isothermals)
Sophie [7]

Answer:

Step-by-step explanation:

Given that:

T(x,y) = \dfrac{100}{1+x^2+y^2}

This implies that the level curves of a function(f) of two variables relates with the curves with equation f(x,y) = c

here c is the constant.

c = \dfrac{100}{1+x^2+2y^2} \ \ \--- (1)

By cross multiply

c({1+x^2+2y^2}) = 100

1+x^2+2y^2 = \dfrac{100}{c}

x^2+2y^2 = \dfrac{100}{c} - 1 \ \  -- (2)

From (2); let assume that the values of c > 0 likewise c < 100, then the interval can be expressed as 0 < c <100.

Now,

\dfrac{(x)^2}{\dfrac{100}{c}-1 } + \dfrac{(y)^2}{\dfrac{50}{c}-\dfrac{1}{2} }=1

This is the equation for the  family of the eclipses centred at (0,0) is :

\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1

a^2 = \dfrac{100}{c} -1  \ \ and \ \ b^2 = \dfrac{50}{c}- \dfrac{1}{2}

Therefore; the level of the curves are all the eclipses with the major axis:

a =  \sqrt{\dfrac{100 }{c}-1}  and a minor axis b =  \sqrt{\dfrac{50 }{c}-\dfrac{1}{2}}  which satisfies the values for which 0< c < 100.

The sketch of the level curves can be see in the attached image below.

7 0
3 years ago
-z=80 solve with steps
dalvyx [7]
-z = 80

You cannot submit an equation with a negative letter ( -z ) so we have to cancel the negative.

To cancel negatives in the equation, we multiply both sides by -1.

-z * -1 = 80 * -1

z = -80

Answer: z = -80
5 0
3 years ago
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