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Aneli [31]
2 years ago
14

Find the unknown length x in the right triangle, to the nearest tenth .

Mathematics
1 answer:
VARVARA [1.3K]2 years ago
8 0

Answer:

x = 11.7

Step-by-step explanation:

sin of an angle is opposite/hypotenuse

sin 38 = x/19

x = 19 sin 38

x = 11.7

You need to memorize the definitions of all the trig functions and then this type of problem will be easy for you.

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5 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
PLEASE NEEED HEELPP
krok68 [10]
There are many systems of equation that will satisfy the requirement for Part A.
an example is y≤(1/4)x-3 and y≥(-1/2)x-6
y≥(-1/2)x-6 goes through the point (0,-6) and (-2, -5), the shaded area is above the line. all the points fall in the shaded area, but
y≤(1/4)x-3 goes through the points (0,-3) and (4,-2), the shaded area is below the line, only A and E are in the shaded area. 
only A and E satisfy both inequality, in the overlapping shaded area.

 
Part B. to verify, put the coordinates of A (-3,-4) and E(5,-4) in both inequalities to see if they will make the inequalities true. 
 for y≤(1/4)x-3: -4≤(1/4)(-3)-3
-4≤-3&3/4 This is valid.
For y≥(-1/2)x-6: -4≥(-1/2)(-3)-6
-4≥-4&1/3 this is valid as well. So Yes, A satisfies both inequalities. 
Do the same for point E (5,-4)

Part C: the line y<-2x+4 is a dotted line going through (0,4) and (-2,0)
the shaded area is below the line
farms A, B, and D are in this shaded area. 
8 0
3 years ago
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