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AfilCa [17]
2 years ago
14

Simify the expression (-11/2 x + 3) -2(-11/4 x -5/2)

Mathematics
1 answer:
Sedaia [141]2 years ago
6 0
Hope this helps a friend

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olaf has 250 files on an external hard drive of those files 12% are photos.(a)write a percent proportion to find the number of f
Ede4ka [16]
A) Olaf has 250 files, which represent 100%. 12% of that are photos which number is unknown (x).
250 files - 100%
x photos - 12%
So, the proportion should look like:
250 files : 100% = x : 12%

B) After crossing the products:
x = 250 files * 12% : 100%
x = 30 photos
8 0
3 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
0.1x+0.2y=-1;0.5x-0.9y=6.4
Harman [31]
.1x+.2y=-1
.5x-.9y=6.4

×-5 equation .1x+.2y=-1

-.5x-y=5
.5x-.9y=6.4

solve by elimination

-1.9y=11.4
÷-1.9 both sides
y=-6

solve for X

.1x+.2 (-6)=-1
.1x-1.2=-1
+1.2 both sides
.1x=.2
÷ 1 both sides
x=2

plug into 2nd equation to see if it's true

x=2
y=-6

0.5(2)-0.9(-6)=6.4
1+5.4=6.4
6.4=6.4 answers hold true
4 0
3 years ago
Solve for y. 8x+6y=10. Please show steps
ddd [48]
8x+6y=10
6y=10-8x
y=(10-8x)/6
6 0
3 years ago
Read 2 more answers
Please help and show work
Artemon [7]

Answer:

175 is the answer

Step-by-step explanation:

5x5=25

25x7=175

6 0
3 years ago
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