1 year ago

Finding the slope passing through the points

Mathematics

1 answer:

marissa [1.9K]1 year ago

3 0

Question 1

Since the x-coordinates are the same, the line is vertical. Therefore, the slope is undefined.

Question 2

Since the y-coordinates are the same, the line is horizontal. Therefore, the slope is zero.

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PLZ

Lena [83]

Answer:

$14.50 I think

cuz u add 6.50+ $8.00=$14.50

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3 years ago

What are three rational numbers between -1 and -2

MArishka [77]

well, rational = fractional, namely something you can write as a fraction, well, anything between -1 and -2 is just less than -1 so

$\bf \boxed{-2}\dotfill \stackrel{-1-\frac{1}{4}}{-\cfrac{5}{4}}\dotfill\stackrel{-1-\frac{1}{10}}{-\cfrac{11}{10}}\dotfill \stackrel{-1-\frac{1}{99}}{-\cfrac{100}{99}}\dotfill \boxed{-1}$

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2 years ago

H. R(x) = kx – 2x^2 ; R(1/4) R(-1/4) find k.

True [87]

Answer:

k=0

Step-by-step explanation:

R(x)=kx-2x²

R(1/4)=k(1/4)-2(1/4)²=k/4-1/8

R(-1/4)=k(-1/4)-2(-1/4)²=-k/4-1/8

R(1/4)=R(-1/4)

k/4-1/8=-k/4-1/8

k/4+k/4=1/8-1/8

2k/4=0

2k=0

k=0

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3 years ago

YALL I need your help! I need to turn this is asap

Gemiola [76]

Answer:

$1.50

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3 years ago

Read 2 more answers

Given the following polynomial find the maximum possible number of turning points. f(x)=x^11+x^10+x^12+x^9

11111nata11111 [884]

Turning points are inflection points

think
1st degree (linear) has no turning/inflecion points
2nd degree (quadratic, parabola) has 1 turning/inflection point
so
nth degree has n-1 turning/inflection point

this is 11th degree since highest power is 12
12-1=11
11 turning oints

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3 years ago