Answer:
9 units2
Step-by-step explanation:
area of a triangle= lengthxwidth/2
3x6=18/2=9
Given parameters:
y varies directly as w;
This interpreted implies that as y increases, we increases by a certain factor or amount;
So, y ∝ w
y = kw
where k is the constant of proportionality
Now, given that y = 8 and w = 2
Input in the equation to solve for k;
8 = k x 2
Solving k gives, k = 4
Now to the second part,
if v is directly proportional to w;
v = k w
So v = 4w
The equation that relates v to w is v = 4w
15/17. The value (ratio) of cos A is 15/17.
The trigonometric ratios of an acute angle are, basically, the sine, the cosine and the tangent. They are defined from an acute angle, α, of a right triangle, whose elements are the hypotenuse, the leg contiguous to the angle, and the leg opposite the angle.
-The sine of the angle is the opposite leg divided by the hypotenuse.
-The cosine of the angle is the adjacent leg divided by the hypotenuse.
-The tangent of the angle is the opposite leg divided by the adjacent leg or, which is the same, the sine of the angle divided by the cosine of the angle.
cos A = adjacent leg/hypothenuse = BC/AC = 15/17
Answer:
Part A) Option A. QR= 3 cm
Part B) Option B. SV=6.5 cm
Step-by-step explanation:
step 1
<u>Find the length of segment QR</u>
we know that
If two triangles are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent
so
In this problem Triangle QRW and Triangle QSV are similar by AA Similarity Theorem
so

we have
---> because S is the midpoint QT (QS=TS)
--->because V is the midpoint QU (QW+WV=VU)
--->because V is the midpoint QU (QV=VU)
substitute the given values

solve for QR

step 2
Find the length side SV
we know that
The <u><em>Mid-segment Theorem</em></u> states that the mid-segment connecting the midpoints of two sides of a triangle is parallel to the third side of the triangle, and the length of this mid-segment is half the length of the third side
so
In this problem
S is the mid-point side QT and V is the mid-point side QU
therefore
SV is parallel to TU
and

so

A. 0.434 B.0.873 C.0.117 D.0.990
A.0.389 B.0.211
Hope I helped
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