Answer:
≈ 78.55 cm
Step-by-step explanation:
Using the ratio of arc (AB) to circumference (C) is equal to the ratio of angle subtended at centre by arc AB to 360°, that is
=
, that is
=
( cross- multiply )
110C = 8640 ( divide both sides by 110 )
C =
≈ 78.55 cm ( to 2 dec. places )
Answer:
<u>It</u><u> </u><u>is</u><u> </u><u>5</u><u>9</u><u>,</u><u>2</u><u>7</u><u>5</u><u>.</u><u>7</u><u>4</u><u> </u><u>grams</u>
Step-by-step explanation:
• Molecular mass of carbon dioxide

[ molar masses: C = 12 g, O = 16 g ]
• From avogadro's number, it says 1 mole contains 6.02 × 10^23 molecules or atoms.
But 1 mole is equivalent to its molar mass in terms of weight, therefore;

Answer:
y=2x
Step-by-step explanation:
y = kx is the format of the graph we are looking for (as told in the instructions) where k is the variable we are supposed to find (like y = x, y=2x, y=3x, etc)
Note that as x increases by 1, y increases by 2. We can see that by writing a table of the points of points that the line given intersects (picture below). We can see that by the values of the x-axis and the y-axis. Why does this happen? This is because y =2x for each value of x. We notice that in our table.
Therefore, the equation is y=2x. I hope this helps! If you still have questions, ask me in the comments!
Answer:
180:270:720
Step-by-step explanation:
total units-> 13u
1 unit -> 1170÷13=90
2u->90×2=180
3u->90×3=270
8u->90×8=720
Answer:
Step-by-step explanation:
The product of distances to the circle along a secant is the same for all secants intersecting a given point.
a. Secants SW and QT intersect at U. Thus ...
(UT)(UQ) = (US)(UW)
(1.5)(4) = (SU)(3)
SU = (4)(1.5)/3
SU = 2
__
b. Secants PT and PX intersect at P. Thus ...
(PQ)(PT) = (PR)(PX)
PX = (PQ)(PT)/PR) = (2.5)(2.5 +4 +1.5)/3 = 20/3
PX = 6 2/3