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stellarik [79]
3 years ago
11

y>mx represent the equation of a line. Where “m” represents the slope of that line. For m> 0, its mean, that the slope of

the line is increasing in a positive direction.
Mathematics
1 answer:
Orlov [11]3 years ago
4 0
I do not know exactly what you are trying to ask, but yes. Any time that the slope is greater than 0, the graph will increase in a positive direction. 
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What expression is equivalent to 8/9 divided 3/4
Semmy [17]

Answer:

32/27 or 1/5/27 or a repeating decimal 1.185

Step-by-step explanation:

4 0
3 years ago
Pls give answer I need to submit today​
BabaBlast [244]

Answer:

a \: 5 \: 10 \: 15 \: 20 \: three \: of \: these \: numbers \\ b \: 2  < 4 < 6 < 8 < 10 < 12 < 14 < 16 <  \\ 18 < 20 \: three \: of \: these \: numbers \\ sum \: of \: numbers < 40 \\  \\ numbers \\ 10 \\ 20 \\ 5 \\ 2

7 0
2 years ago
Is -2 a constant or linear monomial
Allisa [31]
Constant because it’s on its own
6 0
3 years ago
PLEASE HELP I REALLY DONT UNDERSTAND WILL GIVE 100 POINTS
mario62 [17]

Answer:

Part 1) The value of x is 69°

Part 2) Angle 1=64.5°

Part 3) Angle 2=84.5°

Part 4) Angle 3=31°

Part 5) Angle 4=84.5°

Part 6) Angle 5=95.5°

Part 7) Angle 6=95.5°

Part 8) Angle 7=54.5°

Part 9) Angle 8=30°

Step-by-step explanation:

Part 1) Find the value of x

we know that

arc SN+arc QS+arc QP+arc PN=360° -----> by complete circle

substitute the values

60°+x°+(x+40)°+(2x-16)°=360°

solve for x

84°+4x°=360°

4x=276°

x=69°

Part 2) Find the measure of angle 1

we know that

The inscribed angle is half that of the arc it comprises

so

m∠1=(1/2)[arc QSN]

arc QSN=arc QS+SN

arc QSN=x+60°=69°+60°=129°

substitute

m∠1=(1/2)[129°]=64.5°

Part 3) Find the measure of angle 2

we know that

The measure of the inner angle is the semi-sum of the arcs that comprise it and its opposite

m∠2=(1/2)[arc SN+arc QP]

substitute the values

m∠2=(1/2)[60°+(x+40)°]

m∠2=(1/2)[60°+(69+40)°]

m∠2=(1/2)[169°]=84.5°

Part 4) Find the measure of angle 3

we know that

The measurement of the outer angle is the semi-difference of the arcs it encompasses.

m∠3=(1/2)[arc PN-arc SN]

substitute the values

m∠3=(1/2)[(2x-16)°-60°]

m∠3=(1/2)[(2(69)-16)°-60°]

m∠3=(1/2)[62°]=31°

Part 5) Find the measure of angle 4

we know that

m∠4=m∠2 -----> by vertical angles

so

m∠4=84.5°

Part 6) Find the measure of angle 5

we know that

m∠5+m∠2=180° -----> by supplementary angles

so

m∠5+84.5°=180°

m∠5=180°-84.5°=95.5°

Part 7) Find the measure of angle 6

we know that

m∠6=m∠5 -----> by vertical angles

so

m∠6=95.5°

Part 8) Find the measure of angle 7

we know that

The inscribed angle is half that of the arc it comprises

so

m∠7=(1/2)[arc QP]

arc QP=(x+40)°=(69+40)°=109°

substitute

m∠7=(1/2)[109°]=54.5°

Part 9) Find the measure of angle 8

we know that

The inscribed angle is half that of the arc it comprises

so

m∠8=(1/2)[arc SN]

arc SN=60°

substitute

m∠8=(1/2)[60°]=30°

5 0
3 years ago
Which statement about the asymptotes is true with respect to the graph of this function? f(x) = 3x2 – 3 x2 – 4
pochemuha
There are two vertical asymptotes at x = 2 and x = -2.

It looks like you missed a division sign in your problem to create a fraction that would have asymptotes.

The denominator would factor to (x + 2)(x - 2). Since we can't have zero in the denominator, we will have an asymptote at 2 and -2. This is because those values would make the denominator zero.
4 0
2 years ago
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