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Pachacha [2.7K]
3 years ago
11

Need help here please :)

Mathematics
1 answer:
nalin [4]3 years ago
4 0
1. A.Use law of cosines. cosA=(b^2+c^2-a^2)/(2bc) because A is the included angle between b and c. Plug in A=50 degrees, b=13, c=6. cos50=(13^2+6^2-a^2)/(2*13*6), a^2=104.7,a=10.2 approximately, so choose A.

2. A.Use law of cosines again. cosC=(a^2+b^2-c^2)/(2ab). C=95 degrees, a=12, b=22. Plug in, cos95=(12^2+22^2-c^2)/(2*12*22), solve the equation, c^2=674, c=26 approximately. The use law of sines to solve for angle A (also works for B), a/sinA=c/sinC, 12/sinA=26/sin95, sinA=0.46, A=arcsin(0.46)=27.6. Choose A.

3. Answer is A. Area=1/2bc*sinA, since A is the included angle between b and c. Plug in b=30, c=14, A=50 degrees, area=1/2*14*30*sin50=160. 87, so the answer is A.

4. D. As long as the sum of any two sides of the triangle is bigger than the third, the triangle exists. 240+121>263, 240+263>121, 263+121>240, so it exists. To use Heron's formula, first find the semiperimeter, (240+263+121)/2=312. A=\sqrt(312*(312-240)*(312-263)*(312-121))=14499.7 approximately, so choose D.

5. 300. The included angle between the two paths is C=49.17+90=139.17 degrees. The lengths of the two paths are a=150, b=170. c is the distance we want. Use the law of cosines, cosC=(a^2+b^2-c^2)/(2ab). Plug in, c^2=89989, c=300 approximately.
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Select ALL the correct answers.
shepuryov [24]

When a quadrilateral is inscribed in a circle, the opposite angles are supplementary

The description of the angles in the quadrilaterals are:

  • b. m∠M = 55°, m∠J = 48°, and m∠L = 132°
  • d. m∠L = 40°, m∠M = 60°, and m∠K = 120°
  • e. m∠K = 72°, m∠L = 44°, and m∠M = 108°
  • f. m∠J = 105°, m∠K = 65°, and m∠L = 75°

<h3>How to describe the angles</h3>

The quadrilateral is given as: JKLM

The opposite angles are:

  • Angles J and L
  • Angles K and M

The opposite angles are supplementary

So, we have:

J + L = 180

K + M = 180

Next, we test the options

<u>Option (a)</u>

J + L = 180

125 + 35 = 160

This is not true

<u>Option (b)</u>

J + L = 180

48 + 132 =180

This is true

<u>Option (c)</u>

K + M = 180

12 + 78 = 180

This is not true

<u>Option (d)</u>

K + M = 180

120 + 60 = 180

This is true

<u>Option (e)</u>

K + M = 180

72 + 108= 180

This is true

<u>Option (f)</u>

J + L = 180

105 + 75 = 180

This true

Hence, the description of the angles in the quadrilaterals are (b), (d), (e) and (f)

Read more about inscribed quadrilaterals at:

brainly.com/question/26690979

6 0
2 years ago
Which equation has the same slope as y = 3x - 4?
OleMash [197]

Answer:

A. y = 3x - 8  

Step-by-step explanation:

y = 3x - 4

This equation is in the form

y= mx+b where the slope is m

The slope is 3

We are looking for an equation with slope 3

A. y = 3x - 8   yes  

B. y = -3x - 3   no  -3

C. y = -3x - 9   no -3

D. y = -3x - 15  no -3

8 0
3 years ago
The temperature in a city was recorded over a ten-hour period. The graph below shows the relationship between the temperature an
harkovskaia [24]

Answer:

A

Step-by-step explanation:

From the graph, we can see that the <em>temperature increased initially</em>, then <em>started to decrease</em>. <u>All of these NOT in a constant rate, though</u>.

Roughly, the temperature was increasing from start till around 5.5 hours. Then on the temperature decreased.

So we can eliminate B, C, and D. That leaves us with correct answer as A.

5 0
3 years ago
The difference of Mai's age and 12 is 60
Fynjy0 [20]

so shes 72 i think

hope i'm right

5 0
3 years ago
Read 2 more answers
What is an equation of the line that passes through the point (-6, -7) and is
Alexandra [31]

Answer:

Step-by-step explanation:

Product of slope of perpendicular lines = -1

6x + 5y = 30

Write this equation in y = mx + b  form

        5y = -6x + 30

          y = \frac{-6}{5}x+\frac{30}{5}

          y=\frac{-6}{5}x + 6

Slope of this line m₁ = -6/5

m₁ * m₂ = -1

      m₂ = -1÷m₁ = -1 * \frac{5}{-6}  

m_{2}=\frac{5}{6}     & (-6 , -7)

Equation of the required line: y - y₁ = m (x - x₁)

y - (-7) = \frac{5}{6}(x - [-6])\\\\y + 7 = \frac{5}{6}x + 6 *\frac{5}{6}\\\\y = \frac{5}{6}x +5-7\\\\y=\frac{5}{6}x-2

8 0
3 years ago
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