Answer:
looks hard
Step-by-step explanation:
I believe this just uses the Pythagorean Theorem. A^2+B^2=C^2. So:
15^2+B^2=25^2. Simplified this is,
225+B^2=625
625-225=400
B^2=400
√ B=<span>√ 400
</span>B=20
Hello!
The graph y < -2/3x + 2 has a dotted line. That means that any points on the dotted line or not shaded, is not a solution to this inequality. Since we are given 4 choices, we can substitute those values into the given inequality to see if it is true.
A) (0, 2)
2 < -2/3(0) + 2
2 < 2, this is false.
B) (3, 0)
0 < -2/3(3) + 2
0 < -2 + 2
0 < 0, this is false.
C) (0, 0)
0 < -2/3(0) + 2
0 < 2, this is true.
D). (0, 3)
3 < -2/3(0) + 2
3 < 2, this is false.
To check if choice C) (0, 0) is true, we should look at the given graph.
Since (0, 0) is in the shaded area, and is not graphed on the dotted line, therefore, a solution to this linear inequality is C, (0, 0).
Answer:
Area of trapezium = 4.4132 R²
Step-by-step explanation:
Given, MNPK is a trapezoid
MN = PK and ∠NMK = 65°
OT = R.
⇒ ∠PKM = 65° and also ∠MNP = ∠KPN = x (say).
Now, sum of interior angles in a quadrilateral of 4 sides = 360°.
⇒ x + x + 65° + 65° = 360°
⇒ x = 115°.
Here, NS is a tangent to the circle and ∠NSO = 90°
consider triangle NOS;
line joining O and N bisects the angle ∠MNP
⇒ ∠ONS =
= 57.5°
Now, tan(57.5°) = 
⇒ 1.5697 = 
⇒ SN = 0.637 R
⇒ NP = 2×SN = 2× 0.637 R = 1.274 R
Now, draw a line parallel to ST from N to line MK
let the intersection point be Q.
⇒ NQ = 2R
Consider triangle NQM,
tan(∠NMQ) = 
⇒ tan65° =
⇒ QM =
QM = 0.9326 R .
⇒ MT = MQ + QT
= 0.9326 R + 0.637 R (as QT = SN)
⇒ MT = 1.5696 R
⇒ MK = 2×MT = 2×1.5696 R = 3.1392 R
Now, area of trapezium is (sum of parallel sides/ 2)×(distance between them).
⇒ A = (
) × (ST)
= (
) × 2 R
= 4.4132 R²
⇒ Area of trapezium = 4.4132 R²
D , the answer is "six less than a number b"