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aliya0001 [1]
3 years ago
10

Triangle A B C is shown. Angle B C A is a right angle. The length of hypotenuse A B is 26, the length of A C is 10, and the leng

th of B C is 24. Given right triangle ABC, what is the value of tan(A)? Five-thirteenths Twelve-thirteenths Twelve-fifths Thirteen-twelfths
Mathematics
2 answers:
ExtremeBDS [4]3 years ago
7 0

Answer:

<h2>The right option is twelve-fifths</h2>

Step-by-step explanation:

Given a right angle triangle ABC as shown in the diagram. If ∠BCA = 90°, the hypotenuse AB = 26, AC = 10 and BC = 24.

Using the SOH, CAH, TOA trigonometry identity, SInce we are to find tanA, we will use TOA. According to TOA;

Tan (A) = opp/adj

Taken BC as opposite side since it is facing angle A directly and AC as the adjacent;

tan(A) = BC/AC

tan(A) = 24/10

tan(A) = 12/5

The right option is therefore twelve-fifths

Svet_ta [14]3 years ago
4 0

Answer:

12/5

Step-by-step explanation:

100% correct

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Answer:

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Step-by-step explanation:

because the coefficient is largest

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Simplify (1/2)4. Plz answer fast
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Answer: 2

Step-by-step explanation: 1/2 x 4 = 4/2 = 2/1 = 2.

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2 years ago
Two mathematical questions attached below.
BartSMP [9]

Answer:

See below

Step-by-step explanation:

9.( {m}^{3}  {n}^{5} )^{ \frac{1}{4} }   \\   =  m^{\frac{3}{4}}n^{\frac{5}{4}}\\  \\ 10. \sqrt[5]{ \sqrt[4]{x} }  \\  =  \sqrt[5]{ {x}^{ \frac{1}{4} } }  \\  =  {( {x}^{ \frac{1}{4} } )}^{ \frac{1}{5} }  \\   =  {x}^{ \frac{1}{4}  \times  \frac{1}{5} }  \\  =  {x}^{ \frac{1}{20} }  \\  \\ \sqrt[5]{ \sqrt[3]{ {a}^{2} } }  \\  =  \sqrt[5]{ {a}^{ \frac{2}{3} } }  \\  =  {( {a}^{ \frac{2}{3} } )}^{ \frac{1}{5} }  \\   =  {a}^{ \frac{2}{3}  \times  \frac{1}{5} }  \\  =  {a}^{ \frac{2}{15} }

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2 years ago
100 POINT QUESTION!!!!!!!!!!!!!!!!!!!!!!!!!!!!
babunello [35]

Answer:

Ok I don't know waht you mean but the answer is

Step-by-step explanation:

Answer:

x=3/2±(√11)2

x=1.5+1.65831i

x=1.5−1.65831i

Find the Solution for

2x^22−6x+10=0

using the Quadratic Formula where

a = 2, b = -6, and c = 10

x=(−b±√(b^2−4ac))/2a

x=(−(−6)±√((−6)2−4(2)(10)))2(2)

x=(6±√(36−80))/4

x=(6±√−44)/4

The discriminant b^2−4ac<0

so, there are two complex roots.

Simplify the Radical:

x=(6±2√11 i)/4

x=6/4±(2√11 i)4

Simplify fractions and/or signs:

x=3/2±(√11)2

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x=1.5+1.65831i

x=1.5−1.65831i

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2 years ago
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Savatey [412]

Answer:

Perimeter of rectangle before folded = 56 in

Total area after folding = 156 sq in

Step-by-step explanation:

Rectangle before folded:  l = 16 and w = 12

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Figure after folding:   Area of trapezoid + area of rectangle

Area of trapezoid = h(b_{1} + b_{2})/2 = 6(4 + 16)/2 = 60

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Total area after folding = 60 + 96 = 156 sq in.

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8 0
3 years ago
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