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Anna71 [15]
3 years ago
15

Karen works at Southern Technical College as a math teacher. She will be paid $900 for each credit hour she teaches. During the

course of her first year of teaching, she would teach a total of 50 credit hours. The college expects her to work a minimum of 170 days (any less and her salary would be reduced) and 8 hours each day.
What is the minimum number of hours she is expected to work for the year?
Question 9 options:

A. 170
B. 50
C. 8,500
D. 1,360
Mathematics
1 answer:
Free_Kalibri [48]3 years ago
3 0

Answer:

Option D

1360

Step-by-step explanation:

The minimum stipulated number of days to work is given as 170 days and since each day one works for 8 hours then total hours per year on minimum side will be 170*8=1360 hours

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Two\ lines\ are\ perpendicular\ if\ product\ of\ the\ slopes\ is\ equal\ -1.\\\\k:y=\frac{3}{7}x+2\to the\ skolpe\ m_k=\frac{3}{7}\\\\l:y=\frac{7}{3}x-\frac{14}{3}\to the\ slope\ m_l=\frac{7}{3}\\\\m_k\times m_l=\frac{3}{7}\times\frac{7}{3}=1\neq-1\\\\conclusion:the\ lines\ are\ not\ perpendicular\\\\Two\ lines\ are\ parallel\ if\ the\ slopes\ are\ equal.\\\\m_k=\frac{3}{7};\ m_l=\frac{7}{3}\to m_k\neq m_l\\\\conclusion:the\ lines\ are\ not\ parallel


Answer:\boxed{C-The\ lines\ intersect\ but\ are\ not\ perpendicular.}
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3 years ago
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HELP PLSSSSS I GIVE BRAINLY
dedylja [7]

Answer: 40 unit²

Step-by-step explanation:

A square pyramid is composed of 1 square and 4 congruent triangles

STEP ONE. Find the area of the square.

A=b²

b=base

A=b²=4²=14

STEP TWO. Find the area of the triangle

A=(1/2)bh

b=base

h=height

A=(1/2)bh

A=(1/2)(4)(3)

A=(1/2)(12)

A=6

There are 4 congruent triangles, so 6×4=24

The surface area of whole=24+16=40 unit²

Hope this helps!! :)

Please let me know if you have any questions

7 0
3 years ago
What is the value of 4C9 ?
vekshin1
The answer is 0

The formula is

n!/ r!(n-r)!

4!/ 9!(4-9)!
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2 years ago
The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n
GarryVolchara [31]

Answer:

(-1,1),(4,-2)

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point (x,y)

Distance between points (x_1,y_1),(x_2,y_2) is given by \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]

Put (x,y)=(-1,1)

34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34

which is true. So, (-1,1) can be a vertex

Put (x,y)=(4,-2)

34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34

which is true. So, (4,-2) can be a vertex

Put (x,y)=(1,1)

34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22

which is not true. So, (1,1) cannot be a vertex

Put (x,y)=(2,-2)

34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22

which is not true. So, (2,-2) cannot be a vertex

Put (x,y)=(4,-1)

34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30

which is not true. So, (4,-1) cannot be a vertex

Put (x,y)=(-1,4)

34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70

which is not true. So, (-1,4) cannot be a vertex

So, possible points for the vertex are (-1,1),(4,-2)

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3 years ago
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Which expression is equivalent to (4 + 6i)2?
larisa86 [58]
\bf \textit{recall that }i^2=-1\\\\
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