Take the unknown number as 'y'.
=》33% = 33/100
=》1.45 = 145/100
33/100 × y = 1.45
33/100 × y = 145/100
33y = 145/100 × 100/1
33y = 145
y = 145/33
y = 4.39
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RainbowSalt2222 ☔
Step-by-step explanation:
you did not even try a calculator to find the result ?
64 is 8×8 or 8².
so, sqrt(64) = 8.
similarly for 4 :
4 is 2×2 or 2².
so, sqrt(4) = 2
so,
sqrt(64)/ sqrt(4) = 8 / 2 = 4
so, d. is the right answer.
Answer:
Coordinate Q is (0.8, 0.7)
Step-by-step explanation:
We are told that the coordinates of point Pare (0.6,0.1).
This means that along the x-axis, x = 0.6 and along the y-axis, y = 0.1.
Now, by inspection of the graph, we can see that when we count boxes from the origin to the point P, we have 6 boxes. Thus, each box corresponds to 0.1. So, for point Q, from the origin to that point, on the x-axis, we have 8 boxes. Since one box = 0.1, then the x - value of Coordinate Q is 0.8.
On the y - axis, we see that we have one box from the origin up for the corresponding y-value of coordinate P.
This means that one box is 0.1.
For coordinate Q, we will count 7 boxes. Thus, y-value of coordinate Q is 0.7.
Thus,coordinate Q is (0.8, 0.7)
Answer:
°
Step-by-step explanation:
The law of sines is a property of all triangles that relates the sides and angles of a triangle. This property states the following:
Where side (A) is the side opposite angle (<a), side (B) is the side opposite angle (<b), and side (C) is the property opposite angle (<c).
Substitute each of the sides and respective angles into the formula, and solve for the unknown angle (<x). Please note that a triangle with two congruent sides (referred to as an isosceles triangle) has a property called the base angles theorem. This states that the angles opposite the congruent sides in an isosceles triangle are congruent. Therefore, there can be two (<x)'s in this triangle.
One can shorten the equation so it only holds the parts that will play a role in solving this equation,
Now take the cross product in this equation to simplify it further,
Inverse operations, solve this equation for (x),
