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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

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

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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)

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

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3 years ago

What is the value of 4C9 ?

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The answer is 0

The formula is

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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

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Let C be point $(x,y)$

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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

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

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

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

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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