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Leno4ka [110]
4 years ago
10

Solve this expression (8+12)×(6-3)÷2Show your work

Mathematics
2 answers:
rodikova [14]4 years ago
8 0
(8+12)x(6-3)÷2
(20)x(3)÷2
60÷2
30
gizmo_the_mogwai [7]4 years ago
4 0
8+12* 6-3 /2
   ^      ^
16*     3  / 2
       ^
     48 /2
         ^ 
        24
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strojnjashka [21]
X² + 18x + 79 = 0
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4 0
3 years ago
A student begins a proof of the law of cosines. His work is shown.
Phoenix [80]

The student made an error in <u>step 3</u>. To correct this error, he should define <u>side AY</u> in terms of <u>angle X</u>. <u>Pythagorean theorem</u> relates sides x, y, z, and <u>angle X</u>

<h3>The law of cosine.</h3>

In Trigonometry, law of cosine (cos) is given by this mathematical expression:

cosθ = Opp/Hyp

<u>Where:</u>

  • Opp is the opposite side of a right-angled triangle.
  • Hyp is the hypotenuse of a right-angled triangle.
  • θ is the angle.

In this scenario, we can logically deduce that the student made an error in <u>step 3</u>. To correct this error, he should define <u>side AY</u> in terms of <u>angle X</u>.

After correcting this error, the student can use the <u>Pythagorean theorem</u> to relate sides x, y, z, and <u>angle X</u>.

Read more on law of cosine here: brainly.com/question/27613782

#SPJ1

6 0
2 years ago
Two airplanes leave an airport at the same time, the first headed due north and the second at a bearing of N42^ E . At 2:00PM, t
Leno4ka [110]

Answer:

The two airplanes are about 330miles apart.

Step-by-step explanation:

The diagram interpreting the question has been attached to this response.

As shown in the diagram,

i. the airplanes leave at point C.

ii. at 2.00pm the first and second airplanes are at points A and B respectively, where they are 312miles and 487miles away from the starting point C in directions due north and N42E from the point C.

iii. the points A, B and C form a triangle with sides a, b and c.

To solve for the value of c which is the distance between the two planes at 2.00pm, the cosine rule is used.

c² = a² + b² - 2abcosC            --------------(i)

where;

b  = 312miles

a = 487miles

C = 42°

Substitute these values into equation (i) and solve as follows;

c² = (487)² + (312)² - 2(312)(487)cos(42)

c² = (237169) + (97344) - 303888cos(42)

c² = (237169) + (97344) - 303888(0.7431)

c² = 334513 - 225819.1728

c² = 108693.8272

<em>Take the square root of both sides</em>

√c² = √108693.8272

c = 329.69

c ≅ 330miles

Therefore, the two airplanes are far apart by 330miles

3 0
3 years ago
Solve the equation 2^x=512
erica [24]

Answer:

9

Step-by-step explanation:

we know 2^10 is 1024, so 2^9 is 512.

5 0
3 years ago
Use the algebra tiles to model x – 2.
choli [55]

Sorry for being late but the answer is B

3 0
3 years ago
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