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Pie
3 years ago
14

Math triangle fun please help!

Mathematics
2 answers:
IgorC [24]3 years ago
6 0

Answer:

Step-by-step explanation:

You can use the cosine rule to solve this problem:

cos(∠KLJ) = (KL² + JL² - KJ²)/(2*JL*KL) = 21463/21960 = 0.97737

∠KLJ = cos⁻¹(0.97737) = 12.2°

Allisa [31]3 years ago
3 0

Answer:

Step-by-step explanation:

We would apply the law of Cosines which is expressed as

a² = b² + c² - 2abCosA

Where a,b and c are the length of each side of the triangle and A is the angle corresponding to a. Likening the expression to the given triangle, it becomes

JL² = JK² + KL² - 2(JK × KL)CosK

122² = 39² + 90² - 2(39 × 90)CosK

14884 = 1521 + 8100 - 2(3510)CosK

14884 = 9621 - 7020CosK

7020CosK = 9621 - 14884

7020CosK = - 5263

CosK = - 5263/7020

CosK = - 0.7497

K = Cos^- 1(- 0.7497)

K = 138.6° to the nearest tenth

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Differentiate. F(x) = (x3 - 3)2/3 2x f'(x) - 3 8 O f'(x) = 3 8 2x2 f'(x) 3 VX3 -8 f(x) 3 V3-8​
oee [108]

Answer:

f'(x) =   \frac{2{x}^{2}}{ \sqrt[3]{( {x}^{3}  - 8) } }

Step-by-step explanation:

f(x) = ( {x}^{3}  - 8)^{ \frac{2}{3} }  \\  \\ f'(x) =  \frac{2}{3} ( {x}^{3}  - 8)^{ \frac{2}{3} - 1 } (3 {x}^{2}  - 0) \\  \\ f'(x) =  \frac{2}{3} ( {x}^{3}  - 8)^{ \frac{2 - 3}{3}  }  \times 3 {x}^{2} \\  \\ f'(x) =  2{x}^{2}( {x}^{3}  - 8)^{ \frac{ - 1}{3}  } \\  \\ f'(x) =   \frac{2{x}^{2}}{( {x}^{3}  - 8)^{ \frac{ 1}{3}  } } \\  \\ \huge \red{ \boxed{ f'(x) =   \frac{2{x}^{2}}{ \sqrt[3]{( {x}^{3}  - 8) } } }}

3 0
3 years ago
What are the endpoint coordinates for the mid segment of △PQR that is parallel to PQ?
diamong [38]

Answer:

So the end points of the mid segment are:

S(-3.5,0.5)

T(-1,-0.5)

Step-by-step explanation:

First of all we need to list the co-ordinates of the points of the triangle shown.

P\rightarrow(-3,3)

Q\rightarrow(2,1)

R\rightarrow(-4,-2)

We need to find mid segment of the triangle which is parallel to segment PQ. This would mean we need to find midpoints of segment PR and QR and then join the points to get mid segment.

Midpoint Formula:

(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Midpoint of PR:

S((\frac{-3+(-4)}{2},\frac{3+(-2)}{2})\\\\(\frac{-3-4}{2},\frac{3-2}{2})\\\\(\frac{-7}{2},\frac{1}{2} )

S(-3.5,0.5)

Midpoint of QR:

T(\frac{2+(-4)}{2},\frac{1+(-2)}{2})\\\\(\frac{2-4}{2},\frac{1-2}{2} )\\\\(\frac{-2}{2},\frac{-1}{2}

T(-1,-0.5)

So the end points of the mid segment are:

S(-3.5,0.5)

T(-1,-0.5)

By mid segment theorem we know that the line joining midpoints of two sides of a triangle is parallel to the 3rd side.

∴ We know ST is parallel to PQ  

4 0
3 years ago
Read 2 more answers
I need help i cant figure it out and I am struggling show WORK. NO LINKS please
Allushta [10]

Answer:

t = -14

Step-by-step explanation:

(t + 8)( - 2) =12

(t)(-2) + (8)(-2) = 12

-2t + -16 = 12

(subtract to make -16 positive)

-2t - 16 = 12

+16 +16

(add 16 to both sides. 16 and 12)

-2t = 28

(divide)

-2t = 28

/-2 /-2

t = -14

hope this helped!

3 0
3 years ago
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