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Alenkasestr [34]
3 years ago
9

There are 100 sheets of paper in the printer, and the number of sheets of paper, p, left after ,t, minutes of printing is given

by the function p(t) = -8t + 100. How many minutes would it take the printer to use all 100 sheets of paper?
Mathematics
1 answer:
Mice21 [21]3 years ago
8 0
I think so 12 to 13 minutes will it take to use all sheets..
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Use properties of shapes and angles to find the missing measures.
guapka [62]

Answer:

< 1 = 130°, < 2 =50°, < 3 = 50°, < 4= 30°, < 5 = 70°, < 6 = 50°, < 7= 60°, < 8 = 40°, < 9 = 40°, < 10 = 50°,  < 11 = 50°, < 12 = 130°, < 13 = 140°, < 14 =40° , < 15 = 140°

Step-by-step explanation:

< 4 = 90- 60 = 30° because angle 4 and angle 60° are complementary

< 5 = 70°, because vertical angles are congruent

< 2 = 180-130 = 50 °, because 130 ° and 2 are a linear pair

< 6 = 180-130 = 50 °, because 130 ° and 2 are a linear pair

because < 2 = < 6 and are alternate exterior angles then two lines are parallel

<13 = 140°,  corresponding  angles

< 13 = < 15 = 140°, vertical angles

< 8 = < 9 = < 14 = 180-140 = 40°, linear pair and vertical angles

< 3 = 180-90-40= 50 °, because the sum off all angles in a Δ is 180°

< 3= < 10 = < 11 =50° vertical angles , and corresponding angles

< 7= 180 -70 -50 = 60°, because the sum off all angles in a Δ is 180°

<1 = <12 = 180 -50 = 130°, linear pair and vertical angles

4 0
3 years ago
HELP MATH
aleksley [76]

Answer:

18.84 is the answer

Step-by-step explanation:

Can I please be marked as brainlist

8 0
3 years ago
&lt;, =, or &gt;? -9/2 ? -4 1/2
astra-53 [7]
Less than right? Because it’s a negative?
8 0
3 years ago
Determine whether 6/7 and 30/35 are equivalent
aleksandr82 [10.1K]
It is equivalent I believe...hope  this helps!
3 0
3 years ago
Read 2 more answers
Isotope mass (amu) abundance (%) 1 203.97304 1.390 2 205.97447 24.11 3 206.97590 22.09 4 207.97665 52.41 find the atomic mass of
Nata [24]
Answer:
average atomic mass = 207.2172085 amu

Explanation:
To get the average atomic mass of an element using the abundance of its isotopes, all you have to do is multiply each isotope by its percentage of abundance and then sum up all the products.

For the question, we have:
203.97304 amu has an abundance of <span>1.390% (0.0139)
</span>205.97447 amu has an abundance of 24.11% (0.2411)
206.97590 amu has an abundance of 22.09% (0.2209)
207.97665 amu has an abundance of 52.41 % (0.5241)

The average atomic mass can be calculated as follows:
average atomic mass = 203.97304(0.0139) + 205.97447(0.2411)
                                     + 206.97590(0.2209) + 207.97665(0.5241)
average atomic mass = 207.2172085 amu

Hope this helps :)
7 0
3 years ago
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