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2 years ago

Solve on the interval please help!

Mathematics

1 answer:

Licemer1 [7]2 years ago

7 0

Answer:

Ф = 2pi/3 , 4pi/3

Step-by-step explanation:

1 + cos Ф = 1/2

Subtract 1 from each side

1 + cos Ф -1 = 1/2-1

cos Ф = -1/2

Take the inverse cos of each side

cos^-1 ( cos Ф )= cos^-1(-1/2)

There are two solutions for the inverse cos of (-1/2), 2pi/3 and 4pi/3

Ф = 2pi/3 +2 npi

Ф = 4pi/3 + 2n pi where n is an integer

We are limited to the interval between 0 and 2pi

Ф = 2pi/3 , 4pi/3

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Ignore the erase marks but i have no clue how to do this. please help asap

madam [21]

Since they're vertical angles, they're equal in degree measure

since they're equal, we can make the two expressions equal to each other

5a - 1 = 2a + 20
add 1 to both sides

5a = 2a + 21
subtract 2a from both sides

3a = 21
divide both sides by 3a

a = 7

now, plug the answer you found for a into one of the two expressions

5a - 1 becomes (5*7) - 1, which equals 34

to double check that they're equivalent (since they're vertical angles)

2a + 20 becomes (2*7) + 20, which equals 34

both angles are 34 degrees!

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2 years ago

I need to know the volume

lana66690 [7]

Answer:

105 mm^3

Step-by-step explanation:

The volume formula for a triangular prism is (height x base x length) / 2.

(5 x 6 x 7) / 2 = 105 mm cubed = 105 mm^3

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2 years ago

A street has parallel curbs 40 feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The lengt

andreev551 [17]

Answer:

The distance, in feet, between the strip = 12 feet.

Step-by-step explanation:

From the figure attached in relation with the question, we can deduce that crosswalk is a parallelogram where

CD/AB = CE/AE

CD = 40

CE = 50

AE = 15

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2 years ago

What number has the same value as 40 tens

MariettaO [177]

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2 years ago

Plz show work and solve quick please need help!!!!!

Simora [160]

Surface area is just the area of all these 4 triangles plus the rectangle.

First we can find the area of the rectangle.

$\sf A=lw$

Half of the length is 28 cm, so the full length must be 28 * 2 = 56 cm.

$\sf A=(56)(27)$

$\sf A=1512cm^2$

The base for the left and right triangles are 27. The heights would be the net length minus half the length of the rectangle:

$\sf 82.8-28=54.8\? cm$

Calculate the area:

$\sf A=\dfrac{1}{2}bh$

$\sf A=\dfrac{1}{2}(27)(54.8)$

$\sf A=739.8\?cm^2$

We have two of these triangles.

$\sf 739.8\times 2=1479.6\? cm^2$

Now do the other two pair of triangles. The bases for them are 28 + 28 = 56 cm. The heights would be the net width minus the width of the rectangle:

$\sf 81.2-27=54.2\?cm$

Now find the area:

$\sf A=\dfrac{1}{2}bh$

$\sf A=\dfrac{1}{2}(56)(54.2)$

$\sf A=1517.6\? cm^2$

We have two of these triangles.

$\sf 1517.6\times 2=3035.2\?cm^2$

Add all the areas together:

$\sf 1512+1479.6+3035.2=\boxed{\sf 6026.80\? cm^2}$

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3 years ago