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3 years ago

Please help!! 15 points!!

Mathematics

2 answers:

Shkiper50 [21]3 years ago

6 0

Here's the solution in attachment :

vampirchik [111]3 years ago

3 0

$- 1 = - 3 - 4 \cos(4θ + \frac{2\pi}{3} ) \\ - 1 + 4 \cos(4θ + \frac{2\pi}{3} ) = - 3 \\ 4 \cos(4θ + \frac{2\pi}{3} ) = - 3 + 1 \\ 4 \cos(4θ + \frac{2\pi}{3} ) = - 2 \\ \cos(4θ + \frac{2\pi}{3} ) = - \frac{1}{2} \\ \cos(4θ + \frac{2\pi}{3} ) = - \frac{1}{2} \\ \cos(2\pi - (4θ + \frac{2\pi}{3} )) = - \frac{1}{2} \\ 4θ + \frac{2\pi}{3} = arccos( - \frac{1}{2} ) \\ \cos( \frac{4\pi}{3} - 4θ) = - \frac{1}{2}$

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A quadrilateral has angles

Liono4ka [1.6K]

Answer:

95 degrees

Step-by-step explanation:

The sum of the measures of a quadrilateral is 360 degrees. 360-105-80-80 = 95 degrees

4 0

2 years ago

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Hurry Quick, help me do this problem. I don't understand how to do this so please help! (Problem in photo)

matrenka [14]

Answer:

1) The system of equations is $5x+4y=75$ and $x+y=18$

2) The first number is $3$ and the second number is $15$

Step-by-step explanation:

1) Let be "x" the first number and "y" the second number.

Remember that:

a- The word "times" indicates multiplication.

b- A sum is the result of an addition.

c- "Is" indicates this sign: $=$

Then, the sum of 5 times "x" and 4 times "y" is 75, can written as:

$5x+4y=75$

And "The sum of the two numbers is 18" can written as:

$x+y=18$

Therefore, the System of equations is:

$\left \{ {{5x+4y=75} \atop {x+y=18}} \right.$

2) You can use the Elimination Method to solve it:

- Multiply the second equation by -5, add the equations and then solve for "y":

$\left \{ {{5x+4y=75} \atop {-5x-5y=-90}} \right.\\......................\\-y=-15\\\\y=15$

- Substitute the value of "y" into any original equation and solve for "x":

$x+15=18\\\\x=18-15\\\\x=3$

7 0

3 years ago

HELP EASY ASAP MULTIPLE CHOICE

Rudik [331]

Answer:

It's the first one 18/3 =6

Step-by-step explanation:

This is the only answer that has all integers or whole numbers. Hope this helps!

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3 years ago

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How dose north america achieve 480VAC with 3 pole 120VAC

makvit [3.9K]

Answer:

There are a variety of factors driving the

adoption of 415/240VAC distribution. First,

distributing at these increased voltages

enables overall efficiency to be enhanced

by eliminating the need for a voltage

transformation along the power path. For

example, in a 415V system, the L-N voltage

is 240V, which is compatible with most all IT

loads. By shifting the UPS output distribution

voltage to 415/240 V three-phase (L-L/

L-N), most of the existing servers today can

be powered with 240V single-phase (L-N)

without any intermediate transformation.

Removing the need for a transformer

streamlines the power path and can increase

efficiency approximately 1.5 percent.

The adoption of a 415/240VAC distribution

model also can yield additional savings at

the device-level in the form of increased

efficiency for server power supplies. As

previously mentioned, today’s power

supplies can handle a wide range of incoming

voltages. Based on the actual voltage being

fed into the power supply, efficiencies can

range from as low as 80 percent (at 100V

nominal) to as much as 94 percent (at 240V

nominal) across all manufacturers.

Furthermore, while the typical efficiency

differential between 240V and 208V gained

at the server power supply today is a relatively

minimal, only 0.3 percent across most of

the load range, the efficiency differential

between 240V and 120V servers can be as

much as 2 percent for a given power supply.

It is this value that often gets overstated by

some manufacturers when comparing 415V

distribution to “traditional methods,” as

there are relatively few servers or peripherals

installed today that are operating at 120V.

Step-by-step explanation:

7 0

3 years ago

A middle school campus is shaped like a rectangle with a width of 1/4 mile and a length of 1/2 mile. A map of the campus has a w

igomit [66]

Answer:

A) 1:4

Step-by-step explanation:

#We first calculate the campus' area:

$a=lw\\\\=\frac{1}{4}\times \frac{1}{2}\\=0.125 \ mi^2$

#We then calculate the campus' map area in square ft:

$A=lw\\\\=0.5 \times 1\\\\=0.5 ft^2$

Ratio is a comparison between two dimensions:

$Ratio=\frac{0.125}{0.5}\\\\=0.25\\\\=\frac{1}{4}\\\\=1:4$

Hence, the ratio of the area in square miles of the campus to the area in square feet of the map is 1:4

3 0

3 years ago