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Alenkinab [10]
3 years ago
8

This composite figure is made up of three simpler shapes. What is the area of this figure?

Mathematics
2 answers:
Anon25 [30]3 years ago
7 0

Answer: it’s 116

Step-by-step explanation:

boyakko [2]3 years ago
6 0
Im pretty sure the answer is 116 square cm
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One third the sum of two numbers is 23.If one of the numbers is 29.3,what is the other number​
Mariana [72]
39.7 I think I’m not sure
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In the graphic below, if W and X are right angles and WY = XY
elena-14-01-66 [18.8K]
Answer is HL

YZ = YZ
WY = XY

<X and <W is right angle

<span>if the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle, then the triangles are congruent</span>
7 0
3 years ago
In parallelogram abcd, ab=14, bc=20, m(angle)b=54. Find to the nearest tenth the length of diagonal bd, and find measure angle d
muminat
It is is a parallelogram, hence we have to face sides equal in length and the opposite angles are also the same. From the given above we have:
ab=14 and its opposite side cd=14
bc=20 and its opposite side da=20

Solving for the diagonal measurement bd, we have consecutive angles are equal to 180°
∠A+∠B=180°
∠A=180°-54°
∠A=126° , ∠B=54° ,∠C=126° and ∠D=54°
bd²=ab²+da²-2(ab)(da)cos126°
bd²=14²+20²-2*14*20cos126°
bd=30.42 unit

Solving for the angle dbc, we have
cos dbc=bc²+bd²-cd²/a*bc*bd
cos dbc=20²+30.42²-14²/2*20*30.42
dbc=21.76° 
3 0
3 years ago
Pls help if pls do most important 15 number 20 number 26 number 28 number 30 number 17 number 25 number 22 number 24 number if u
Tamiku [17]

Answer:

Step-by-step explanation:

15. \frac{cotx+1}{cotx-1} = \frac{cotx+cotx.tanx}{cotx-cotx.tanx} = \frac{cotx(1+tanx)}{cotx(1-tanx)} = \frac{1+tanx}{1-tanx}   \\

16.  \frac{1+cos\alpha }{sin\alpha } = \frac{sin\alpha }{1-cos\alpha }

=> (1+cos\alpha )(1-cos\alpha ) = sin^{2} \alpha \\=> 1-cos^{2}\alpha =sin^{2}  \alpha \\=> sin^{2}\alpha +cos^{2}\alpha =1\\

=> The clause is correct

17. Do the same (16)

18. \frac{sinx}{1-cosx}+\frac{sinx}{1+cosx} = \frac{sinx(1+cosx) + sinx(1-cosx)}{(1+cosx)(1-cosx)} = \frac{sinx + sinx.cosx + sinx - sinx.cosx}{1-cos^{2}x} = \frac{2sinx}{sin^{2}x }= \frac{2}{sinx} = 2cosecx

19.

\frac{sinA}{1+cosA} + \frac{1+cosA}{sinA}=\frac{2}{sinA} \\=> \frac{sinA}{1+cosA} + \frac{1+cosA}{sinA} - \frac{2}{sinA} \\ = 0\\=> \frac{sinA}{1+cosA} + (\frac{1+cosA}{sinA} - \frac{2}{sinA} \\) = 0\\=> \frac{sinA}{1+cosA} + \frac{1+cosA-2}{sinA} = 0\\=> \frac{sinA}{1+cosA} +\frac{cosA-1}{sinA} = 0\\=> \frac{sin^{2} A+(cosA-1)(1+cosA)}{(1+cosA)sinA}=0\\=> \frac{sin^{2} A+cos^{2} A-1}{(1+cosA)sinA}=0\\=> \frac{0}{(1+cosA)sinA} =0\\

=> The clause is correct

20. Do the same (18)

21. Left = \frac{1}{cosecA+cotA} = \frac{1}{\frac{1}{sinA}+\frac{cosA}{sinA}  } = \frac{1}{\frac{1+cosA}{sinA} }= \frac{sinA}{1+cosA}

Right = cosecA-cotA = \frac{1}{sinA}-\frac{cosA}{sinA}= \frac{1-cosA}{sinA}   \\

Same with (16) => Left = Right => The clause is correct

22. Do the same (21)

Too long, i'm so lazy :))))

5 0
3 years ago
(2x^3 - 2x^2) Divide or subtract exponents ?
Agata [3.3K]

Answer:

Neither.

Step-by-step explanation:

You can only add exponents when multiple, or subtract when dividing.

Also, 2x^3 and 2x^2 are two different terms, not like terms, so cannot subtract the two.

6 0
3 years ago
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