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2 years ago

Plz help fifty points

Mathematics

2 answers:

Tomtit [17]2 years ago

8 0

Answer:

...

Step-by-step explanation:

likoan [24]2 years ago

7 0

Answer:

for question 13

Step-by-step explanation:

for 14

x. 150

y. 15

z. 150

i cant say that this is totally correct but im taking a educated guess

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A wire that is 0.20 meters long is moved perpendicularly through a magnetic field of strength 0.45 newtons/amperes meter at a sp

Natali [406]

Answer:

Given:

length of the wire = 0.20 meters

magnetic field strength = 0.45 newtons/amperes meter

speed = 10.0 meters per second

emf = B * l * v

B = flux density ; l = length of the wire ; v = velocity of the conductor

emf = 0.45 newtons / ampere meter * 0.20 meters * 10.0 meters/seconds

emf = 0.90 volts

The emf produced is 0.90 volts.

Step-by-step explanation:

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2 years ago

Patrick buys 12 bunches of bananas for $9 what will he pay for 8 bunches of bananas

larisa [96]

$12b.......9\$ \\ 8b.........x \\\\ x=\frac{8*9}{12}=\frac{72}{12} \\\\ \boxed{x=5\$}$

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3 years ago

What is the value of x? Leave answer in simplest radical form.

solniwko [45]

Hypotenuse or C (x) can be found with the equation $\sqrt{6 ^{2} + 2 ^{2} }$
After solving you get about 6.32

6.32 about the same as $2 \sqrt{10}$

4 0

3 years ago

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Solve for c of the right triangle

Ede4ka [16]

In order to solve this triangle we must use cosine law:
$c^{2} = a^{2} + b^{2} - 2ab*cosC$

Now let's plug in the values we are given:
$c^{2} = 5^{2} + 8^{2} - 2(5)(8) * cos 67$
$c^{2} = 25 + 64 - 80*cos67$
$c = \sqrt{80-cos67}$

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3 years ago

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a tight roll of paper, as delivered to the printer, is 60cm in diameter, and the paper is wound on to a wooden cylinder 8cm in d

Alekssandra [29.7K]

The area of the paper is the amount of space it occupies.

The length of papers around the edge is 1607680 cm

The given parameters are:

$\mathbf{d_1 = 60cm}$

$\mathbf{d_2 = 60cm + 8cm = 68cm}$

The area of the circle head is:

$\mathbf{Area = \pi \frac{d^2}{4}}$

So, we have:

$\mathbf{Area = \pi \frac{d_2^2 - d_1^2}{4}}$

Substitute known values

$\mathbf{Area = \pi \frac{68^2 - 60^2}{4}}$

$\mathbf{Area = \pi \frac{1024}{4}}$

$\mathbf{Area = 256\pi}$

The length of the paper is then calculated as:

$\mathbf{Length = \frac{Area}{0.0005}}$

This gives

$\mathbf{Length = \frac{256\pi}{0.0005}}$

$\mathbf{Length = \frac{256 \times 3.14}{0.0005}}$

$\mathbf{Length = 1607680}$

Hence, the length of papers around the edge is 1607680 cm