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3 years ago

Simplify the expression by using a double-angle formula. 2 sin 31°cos 31°

Mathematics

1 answer:

qwelly [4]3 years ago

6 0

The double angle formula for the sine is

$\sin(2x) = 2\sin(x)\cos(x)$

The right hand side is exactly your expression, where $x = 31^\circ$

So, rewriting the expression from right to left, we have

$2\sin(31^\circ)\cos(31^\circ) = \sin(2\cdot 31^\circ) = \sin(61^\circ)$

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Is the line through points P(0, 1) and Q(5, 0) parallel to the line through points R(–4, 3) and S(–5, 4)? Explain.

ale4655 [162]

Answer:

Step-by-step explanation:

Let's find the slope of both lines and then compare these slopes:

From (-4,3) to (-5,4) entails a decrease of -1 in x and an increase of 1 in y. Thus, the slope is m = rise / run = 1/(-1) = -1.

From (0,1) to (5,0) entails an increase of 5 in x and a decrease of 1 in y. The slope here is m = rise / run = -1/5.

For these lines to be parallel, their slopes must be the same. That is not the case here, so NO, the lines are not parallel.

8 0

3 years ago

Use the given equation to determine if the points in the brackets are on the graph

dmitriy555 [2]

The points in the brackets are on the graph are (0, 2) and (6, 10)

<h3>How to determine if the points in the brackets are on the graph?</h3>

The equation of the line is given as

4x - 3y = -6

The points are given as:

{(0,2),(1,3),(4,7),(6,10)}

Rewrite as

(x, y) = {(0,2),(1,3),(4,7),(6,10)}

Next, we substitute the x and y values in the equation 4x - 3y = -6

So, we have

(0, 2):

4(0) - 3(2) = -6

-6 = -6 ---- true

(1, 3):

4(1) - 3(3) = -6

-5 = -6 ---- false

(4, 7):

4(4) - 3(7) = -6

-5 = -6 ---- false

(6, 10):

4(6) - 3(10) = -6

-6 = -6 ---- true

Hence, the points in the brackets are on the graph are (0, 2) and (6, 10)

Read more about linear equations at:

brainly.com/question/2226590

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3 0

1 year ago

After the fine arts booster club's soup supper, there were 120 pieces of pie left over. The leftover pie was distributed equally

Ne4ueva [31]

Answer: Your answer is 12

Step-by-step explanation:

120/(6+4)=X

120/10=X

120/10=12

12=X

8 0

3 years ago

Algebra 2 semester 2 activity 4.1.1

podryga [215]

I don’t know what your asking

8 0

3 years ago

The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base and PA = 2sqrt5 in. The base edges AB = 10 in, AC

DIA [1.3K]

Answer:

PQ = √84 = 2√21 in ≈ 9.165 in

Step-by-step explanation:

The base edges AB = 10 in, AC = 17 in, and BC = 21 in

Q ∈ BC and AQ is the altitude of the base.

Let BQ = x in ⇒ CQ = (21 - x) in

ΔAQB a right triangle at Q

∴ AQ² = AB² - BQ² = 10² - x² ⇒(1)

ΔAQC a right triangle at Q

∴ AQ² = AC² - CQ² = 17² - (21-x)² ⇒(2)

Equating (1) and (2)

∴ 10² - x² = 17² - (21-x)²

∴ 10² - x² = 17² - (21² - 42x + x²)

∴ 10² - x² = 17² - 21² + 42x - x²

∴ 10² - 17² + 21² = 42x

∴ 42x = 252

∴ x = 252/42 = 6

Substitute at (1)

∴ AQ² = AB² - BQ² = 10² - x² = 100 - 36 = 64 = 8²

∴ AQ = 8 in

∵ The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base

∴ PA⊥AQ

∴ ΔPAQ is a right triangle at A,

PA = 2sqrt5 in and AQ = 8 in

∴ PQ (hypotenuse) = √(PA² + AQ²)

∴ PQ = √[(2√5)² + 8²] = √(20+64) = √84 = 2√21 in ≈ 9.165 in

6 0

3 years ago