650 kgs is equal to 1,433 lbs. I believe the mistake would be that they forgot to add a period in between 65 and the 0, because 65.0 kgs is equal to 143.3 lbs.
Answer:
25
Step-by-step explanation:
When we write expressions for the total cost of each field visit and set them equal, we find the solution to be the ratio of the difference in fixed cost to the difference in variable cost.
y = 75 +7x . . . . . cost for x students to visit the science center
y = 50 +8x . . . . cost for x students to visit the natural history museum
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Subtracting the first equation from the second, we get ...
0 = -25 +x
25 = x . . . . . add 25; the number of students such that costs are equal
The cost will be the same either place for 25 students.
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<em>Additional comment</em>
Here, the fixed cost difference is 75-50=25, and the variable cost difference is 8-7=1. The ratio of these costs is ...
$25/($1 /student) = 25 students.
This relationship only holds when the higher fixed cost is associated with the lower variable cost. Charges are such that one place caters to larger numbers of students (science center), and one prefers fewer students (natural history museum).
<span>A’(2, -4); B’(7, -9); C’(-2, -9)
I believe you did it correctly but over the wrong line. You were supposed to do it over line M and it looks like you did it over the X-axis.</span>
Answer:
54
Step-by-step explanation:
To solve problems like this, always recall the "Two-Tangent theorem", which states that two tangents of a circle are congruent if they meet at an external point outside the circle.
The perimeter of the given triangle = IK + KM + MI
IK = IJ + JK = 13
KM = KL + LM = ?
MI = MN + NI ?
Let's find the length of each tangents.
NI = IJ = 5 (tangents from external point I)
JK = IK - IJ = 13 - 5 = 8
JK = KL = 8 (Tangents from external point K)
LM = MN = 14 (Tangents from external point M)
Thus,
IK = IJ + JK = 5 + 8 = 13
KM = KL + LM = 8 + 14 = 22
MI = MN + NI = 14 + 5 = 19
Perimeter = IK + KM + MI = 13 + 22 + 19 = 54
Answer:
4.58
Step-by-step explanation:
21 is not a perfect square, but lies between the perfect squares 16 and 25. Thus, √21 lies between √16 and √25, that is, between 4 and 5. Since 21 is closer to 25 than to 16, √21 is closer to 5 than to 4. A rough estimate of √21 would be 4.7.
√21 done on a calculator comes out to 4.58.
