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NISA [10]
3 years ago
10

Please help me there is a picture right there

Mathematics
2 answers:
CaHeK987 [17]3 years ago
8 0

Answer:

90 mm^{2}

Step-by-step explanation:

Hi!

What we have over here is a composite figure.

Basically, it's made up of different shapes

We can either make a rectangle by making a horizontal line from the point in between 6 mm and 5 mm

Or we can make a rectangle by going verticle from the point in between        6 mm and 5 mm

Either way, we have two rectangles now, and we can use our formula l*w to find the area.

For the first example, we would do:

6*5 =30

6*10=60

Then we add the two areas of the two rectangles:

60+30=90 mm^{2}

For the second example, we would do:

5*6=30

5*12=60

60+30=90 mm^{2}

Just cut the figure into shapes you can calculate the area of and you should be good to go!

Let me know if you have any questions

seraphim [82]3 years ago
3 0

Answer:

90 mm^2

Step-by-step explanation:

split into two rectangles.

First rectangle: 5 x 6 = 30 mm^2

Second rectangle: 10 x 6 = 60 mm^2

30 + 60 = 90 mm^2

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6 0
3 years ago
30 points pls helppp
Phoenix [80]

Answer:

(y + 1) = 3( x +2)

Step-by-step explanation:

from equation,

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given x1=2, y1= -1

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substitute

y - (-1) = 3( x - 2)

ans: y+1= 3(x+2)

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2 years ago
Please help! I’ll mark brainliest
Usimov [2.4K]

Answer: 19 ≥ 3z + 1 ≥ - 5

8 0
3 years ago
Helppppppp pleaseee
Viktor [21]

Answer:

x=2\sqrt{5}

Step-by-step explanation:

If that is a square, then all sides are the same. Therefore, it's a \sqrt{10} by \sqrt{10} shape. We'll use Pythagorean Theorem to find the diagonal.

a^2+b^2=c^2\\\sqrt{10}^2+ \sqrt{10}^2=x^2\\10+10=x^2\\20=x^2\\\sqrt{20}=x\\\sqrt{4*5}=x\\ 2\sqrt{5}=x

Since it's not a fraction, there is no denominator to rationalize. Let me know if you have any other questions!

5 0
2 years ago
Can some show the steps and the answer please I really need help tell me how you get it and the answer
Lana71 [14]

Answer:

Step-by-step explanation:

so this is about triangles..  sooo the following is a bit of helpful reminders that I keep on my computer to help me remember how to fit the trig functions to triangles..  I strongly suggest you copy it and keep it where you can look at it often.

Use SOH CAH TOA to recall how the trig functions fit on a triangle

SOH: Sin(Ф)= Opp / Hyp

CAH: Cos(Ф)= Adj / Hyp

TOA: Tan(Ф) = Opp / Adj

I use this anytime I run into triangles or need some help with sin or cos

now the problem , 3 ladder  10, 12, & 15 feet.   Alex wants to get to 8 feet.

the problems is also telling you that   you can use   t.....   and then , the words are cut off.. but I know they were going to say   Tan ... next.. :P

b/c Tan is how you figure out problems with the adjacent side and the opposite side.  like this problem.    Look at  TOA  above.  use that to recall how the parts fit in the formula

Tan(∅) = Opp / Adj

they give us the Opp side of 8 feet in the problem

then they also tell us the Hyp of the triangle which is each of the ladders length.  Then they ask us what is the Adj sides length?

So we also need to solve the triangle with the know hyp  (ladder length).. uggg, this problem is long. Then we can solve the dist. from the wall or Adj side length.

it's two steps, if you want to think of it that way.   You're supposed to be pretty confident with trig functions.   I'm guessing this is a trig class.. right?

let's solve for the 3 different angles that the ladders make , each going to 8 feel.  Obviously, nobody would really do this with a ladder they would just lean it against the wall . and if it's taller than where they want to climb, they would just go up part way.   so anyway,     find the 3 different angles.  

look above to see which formula to use.

I like SOH b/c it seems to have all the pieces of the triangle we want to work with.

ladder 1  ( 10')

Sin(∅) = Opp / Hyp

Sin(∅) = 8 / 10

∅ = arcSin (4/5)

[ first, yes, I just reduced the fraction, then I did the arcSin on both sides, I think you might know how to do that already ? ]

∅ = 53.13010 °

( yes, I used my calculator to find that,  calculators are okay to use when figuring out non standard angles )

ladder 2 (12')

Sin(∅) = 8/12

∅ = arcSin (2/3)

∅ = 41.81031°

ladder 3 (15')

Sin(∅) = 8/15

∅ = arcSin (8/15)

∅ = 32.230952°

now use our Tan function to find the Adjacent side which is the distance from the wall

Tan(∅)= Opp / Adj

Adj = Opp / Tan(∅)

( I did some quick algebra to move the side we want to solve for, now plug and chug all 3 angles  )

ladder 1

Adj = 8 / Tan(53.13010)

Adj = 6.0000005    

ladder 2

Adj = 8 / Tan(41.81031)

Adj = 8.94427

ladder 3

Adj = 8 / Tan(32.230952)

Adj = 12.688577

so the 10' ladder is 6 feet from the wall

the 12' ladder is 8.9 feet from the wall

the 15 foot ladder is 12.7 feet from the wall.

I really don't think that 15' ladder is going to stay on the wall.. if Alex climbs it... it's way way too far out... it will just fall straight down the wall  :/   Maybe another math problem for the forces involved  :P

5 0
3 years ago
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