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Gala2k [10]
3 years ago
7

How would I solve this ?

Mathematics
1 answer:
katrin2010 [14]3 years ago
5 0

Answer:

Step-by-step explanation:

7/x=tan32

x=7/ tan 32

or x=11.20

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The model below can be used to find the quotient of one over two divided by one over six. What is the quotient?
VLD [36.1K]

Answer:

i like it is 1/2 can you make smaller hope this help

Step-by-step explanation:

3 0
2 years ago
Write the equation of a line that is
zvonat [6]

Answer:

y=0.5x-5

Step-by-step explanation:

Since the equation is parallel, this means that the slope remains the same which is 0.5.

The remaining thing to do is to find b or y-intercept. To do that, the point (6,-2) and the slope of 0.5 will be used.

-2=0.5(6)+b

When solving for b, it will equal to -5.

Therefore the new equation that is parallel is y=0.5x-5

6 0
3 years ago
Divide x^3-3x^2-10x+24 by x-1
lesya [120]

Answer:

x^2 - 2x - 12 with remainder 12

Step-by-step explanation:

Synthetic division is the fastest way in which to carry out this division.

The divisor (x - 1) from long division corresponds to the divisor 1 in synthetic division.  Setting up synthetic division, we get:

1     /      1     -3     -10     24

                     1      -2     -12

      --------------------------------

             1      -2     -12      12

The first three digits {1, -2, -12} are the coefficients of the quotient, and 12 represents the remainder:

The quotient is 1x^2 - 2x - 12 and the remainder is 12.

6 0
3 years ago
What’s the answer to number 18 please?!
kozerog [31]

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4 0
3 years ago
2. A solar lease customer built up an excess of 6,500 kilowatt hours (kwh) during the summer using his solar
tia_tia [17]

9514 1404 393

Answer:

  a)  E = 6500 -50d

  b)  5000 kWh

  c)  the excess will last only 130 days, not enough for 5 months

Step-by-step explanation:

<u>Given</u>:

  starting excess (E): 6500 kWh

  usage: 50 kWh/day (d)

<u>Find</u>:

  a) E(d)

  b) E(30)

  c) E(150)

<u>Solution</u>:

a) The exces is linearly decreasing with the number of days, so we have ...

  E(d) = 6500 -50d

__

b) After 30 days, the excess remaining is ...

  E(30) = 6500 -50(30) = 5000 . . . . kWh after 30 days

__

c) After 150 days, the excess remaining would be ...

  E(150) = 6500 -50(150) = 6500 -7500 = -1000 . . . . 150 days is beyond the capacity of the system

The supply is not enough to last for 5 months.

3 0
2 years ago
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