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saveliy_v [14]
3 years ago
7

What is m Angle Q B O? 5° 9° 12° 24°

Mathematics
1 answer:
anzhelika [568]3 years ago
5 0
12 cant be 5 9 or 24 so go with 12
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A framer's market sells apples for $0.75 per pound and oranges for $0.89 per pound
kramer

Answer:

160

Step-by-step explanation:

jbuyvcezxrctvubnjonjbvcrexwexcrvbnmnbvcrxewexrcvtbn

8 0
2 years ago
What is 5au + 24av - 5bu - 30bv​
Blizzard [7]

Answer:

59

Step-by-step explanation:

5au+24av-5bu-30bv

rearrange it

5au+5bu-24av-30bv

similar letter cross each other

5a+5b-24a-30b

rearrange it

5a+24a-5b-30b

similar letter cross each other

5+24-5-30

29- -30=59

-and - = +

5 0
2 years ago
Write an equation in point slope form for the line with a slope of 3 passes through (1,2)
alexgriva [62]

Answer:

y-2=3(x-1)

Step-by-step explanation:

The point-slope form of a line is given as:

y-y_1=m(x-x_1)

Where

m is the slope

x_1 is the x-coordinate of the point given (passing through the line)

y_1 is the y-coordinate of the point given (passing through the line)

Now,

Given,

Slope = m = 3

x_1 = 1

y_1 = 2

Now, we simply plug these into the formula for point-slope form of a line:

y-y_1=m(x-x_1)\\y-2=3(x-1)

This is the point-slope form.

4 0
3 years ago
guys i have a question me and my girlfriend are still in love but we live in to different zones of louisiana.I want to suprise h
notka56 [123]

Answer:

this is my opinion so

Step-by-step explanation:

Well congrats : and suprise her with stuff she likes , Flowers, candy,cloths but her favorites of each thing. honestly anything she will still like it either way :) i hope i helpee

3 0
3 years ago
Prove the trigonometric identity
Annette [7]

Answer:

Proved See below

Step-by-step explanation:

Man this one is a world of its own :D Just a quick question are you a fellow Add Math student in O levels i remember this question from back in the day :D Anyhow Lets get started

For this question we need to know the following identities:

1+tan^{2}x=sec^2x\\\\1+cot^2x=cosec^2x\\\\sin^2x+cos^2x=1

Lets solve the bottom most part first:

1-\frac{1}{1-sec^2x} \\\\

Take LCM

1-\frac{1}{1-sec^2x} \\\\\frac{1-sec^2x-1}{1-sec^2x} \\\\\frac{-sec^2x}{1-sec^2x} \\\\\frac{-(1+tan^2x)}{-tan^2x}

now break the LCM

\frac{-1}{-tan^2x}+\frac{-tan^2x}{-tan^2x}\\\\\frac{1}{tan^2x}+1\\\\cot^2x+1

because 1/tan = cot x

and furthermore,

cot^2x+1\\cosec^2x

now we solve the above part and replace the bottom most part that we solved with cosec^2x

\frac{1}{1-\frac{1}{cosec^2x} } \\\\\frac{1}{1-sin^2x} \\\\\frac{1}{cos^2x}\\\\sec^2x

Hence proved! :D

4 0
3 years ago
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