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zaharov [31]
3 years ago
9

Find the tangents of the acute angles in the right triangle

Mathematics
1 answer:
crimeas [40]3 years ago
8 0

f=50

d=40

Hope this helps

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Help with this word problem???? :)
ycow [4]

Answer:

I'm pretty sure it's 144

Step-by-step explanation:

1.           160 divided by 5 is 32 so 1 quart = 32 pounds

2.          32 times 4.5 = 144

P.S Mark as brainliest rate 5 stars and hit that thanks button!

4 0
3 years ago
Choose the three equivalent forms of 4.625.
Ede4ka [16]

Answer:

Step-by-step explanation:

4.625%- percents and decimals are the same

four and five eighths- five divided by eight is (.625) and 4+ (.625) is 4.625

thirty seven eighths- 37 divided by 8 is 4.625

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1 year ago
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Asma was asked to compare the following two numbers.
GalinKa [24]

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No

Step-by-step explanation:

6.212 x 10^8 = 621,200,000

4.73  x 10^9 = 4,730,000,000

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2 years ago
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Here are the first five terms of a different quadratic sequence.
Elina [12.6K]

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4 0
2 years ago
Suppose that 5 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 51 cm. (a) How much work
olganol [36]

Answer:

a) 0.6 joules of work are needed to stretch the spring from 41 centimeters to 46 centimeters.

b) The spring must be 5.6 centimeters far from its natural length.

Step-by-step explanation:

a) The work done to stretch the ideal spring from its natural length is defined by the following definition:

W = \frac{1}{2}\cdot k\cdot (x_{f}-x_{o})^{2} (1)

Where:

k - Spring constant, measured in newtons.

x_{o}, x_{f} - Initial and final lengths of the spring, measured in meters.

W - Work, measured in joules.

The spring constant is: (W = 5\,J, x_{o} = 0.36\,m, x_{f} = 0.51\,m)

k = \frac{2\cdot W}{(x_{f}-x_{o})^{2}}

k = \frac{2\cdot (5\,J)}{(0.51\,m-0.36\,m)^{2}}

k = 444.44\,\frac{N}{m}

If we know that k = 444.44\,\frac{N}{m}, x_{o} = 0.41\,m and x_{f} = 0.46\,m, then the work needed is:

W = \frac{1}{2}\cdot \left(444.44\,\frac{N}{m} \right)\cdot (0.46\,m-0.41\,m)^{2}

W = 0.555\,J

0.6 joules of work are needed to stretch the spring from 41 centimeters to 46 centimeters.

b) The elastic force of the ideal spring (F), measured in newtons, is defined by the following formula:

F = k\cdot \Delta x (2)

Where \Delta x is the linear difference from natural length, measured in meters.

If we know that k = 444.44\,\frac{N}{m} and F = 25\,N, then the linear difference is:

\Delta x = \frac{F}{k}

\Delta x = \frac{25\,N}{444.44\,\frac{N}{m} }

\Delta x = 0.056\,m

The spring must be 5.6 centimeters far from its natural length.

3 0
3 years ago
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