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matrenka [14]
3 years ago
6

Three lines intersect to form 6 angles if the measure of <2 is 53° what can be concluded ?

Mathematics
1 answer:
lys-0071 [83]3 years ago
5 0

Answer:  '

Step-by-step explanation:

'

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2) Simplify: a(0.5 + 8)
kipiarov [429]

Answer:

(8.5)

Step-by-step explanation:

Just add the 2 numbers

4 0
3 years ago
Someone help I don't know this trig question ​
solong [7]

Answer:

Bill is 9.80 km apart from Concert

Step-by-step explanation:

It's given in the question,

m∠B = 75°

m∠C = 45°

By triangle sum theorem,

m∠A + m∠B + m∠C = 180°

m∠A + 75° + 45° = 180°

m∠A = 180° - 120°

m∠A = 60°

By sine rule,

\frac{\text{sin}(\angle A)}{BC}=\frac{\text{sin}(\angle C)}{AB}

\frac{\text{sin}(60)}{BC}=\frac{\text{sin}(45)}{8}

BC = \frac{8\text{sin}(60)}{\text{sin}(45)}

     = 9.80 km

Therefore, Bill is 9.80 km apart from Concert.

8 0
3 years ago
Identify the number of decimal places in each factor. 0.47 × 3.2 The number of decimal places in 0.47 is . The number of decimal
nignag [31]

Answer:

2, 1, 3

Step-by-step explanation:

correct on edg

5 0
4 years ago
Urgent!! Will mark brainliest!!
horsena [70]

Answer:

1) x is negative and y is positive ⇒ last answer

2) cotФ = -12/35 ⇒ second answer

3) The right identity is cot²Ф - csc²Ф = -1 ⇒ last answer

Step-by-step explanation:

* For any point (x , y) lies on the terminal side of the angle Ф

 in standard position

* x = cosФ and y = sinФ

- If Ф in the first quadrant, then x , y are positive

∴ All trigonometry functions are positive

- If Ф in the second quadrant, then x is negative , y is positive

∴ sinФ only is positive

- If Ф in the third quadrant, then x is negative , y is negative

∴ tanФ only is positive

- If Ф in the fourth quadrant, then x is positive , y is negative

∴ cosФ only is positive

* Lets solve the problems

∵ Ф = 3π/4 ⇒ (135°)

∴ It lies on the second quadrant

∴ x is negative and y is positive

* Lets revise the reciprocal of sinФ, cosФ and tanФ

- cscФ = 1/sinФ

- secФ = 1/cosФ

- cotФ = 1/tanФ

∵ secФ = -37/12

∴ cosФ = -12/37

∵ π/2 < Ф < π

∴ Ф lies on the second quadrant

∴ cotФ is negative values

∵ tan²Ф = sec²Ф - 1

∵ secФ = -37/12

∴ tan²Ф = (-37/12)² - 1 = 1225/144 ⇒ take√ for both sides

∴ tanФ = ± 35/12

∵ cotФ = ± 12/35

∵ cotФ is negative value

∴ cotФ = -12/35

* In the standard position of the angle Ф the terminal

 of it lies on the unit circle O

- By using Pythagorean theorem

∵ x² + y² = 1

∵ x = cosФ and y = sinФ

∴ cos²Ф + sin²Ф = 1 ⇒ (1)

∴ cos²Ф = 1 - sin²Ф

∴ sin²Ф = 1 - cos²Ф

* Divide (1) by cos²Ф

∴ cos²Ф/cos²Ф + sin²Ф/cos²Ф = 1/cos²Ф

* Remember sin²Ф/cos²Ф = tan²Ф and 1/cos²Ф = sec²Ф

∴ 1 + tan²Ф = sec²Ф ⇒ (2) ⇒ subtract 1 from both sides

∴ tan²Ф = sec²Ф - 1 ⇒ subtract sec²Ф from both sides

∴ tan²Ф - sec²Ф = -1

* Divide (1) by sin²Ф

∴ cos²Ф/sin²Ф + sin²Ф/si²Ф = 1/sin²Ф

* Remember cos²Ф/sin²Ф = cot²Ф and 1/sin²Ф = csc²Ф

∴ cot²Ф + 1 = csc²Ф ⇒ (3) ⇒ subtract 1 from both sides

∴ cot²Ф = csc²Ф - 1 ⇒ subtract csc²Ф from both sides

∴ cot²Ф - csc²Ф = -1

* The right identity is cot²Ф - csc²Ф = -1

3 0
4 years ago
60 Point Question Answer ASAP! Brainliest to best/first answer
MAVERICK [17]

Step-by-step explanation:

the change rate between the data points gives us the print rate.

so,

(f(x2) - f(x1)) / (x2 - x1)

starting with x1 = 10 and x2 = 15 we see

(133 - 98) / (15 - 10) = 35/5 = 7

that means he is printing 7 pages per minute.

we see that we have the same difference (35 prints) for every following 5-minute interval.

so, that is the confirmed print rate.

and yes, the amount is increasing, of course.

after 10 minutes he has 98 flyers but we know he could have printed only 7×10 = 70 flyers in that time.

so, he started with 98-70 = 28 flyers before actually printing new ones.

4 0
3 years ago
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