Answer:
123
Step-by-step explanation:
because if its lower than 50 you round down, if its 50 or above you round up
The <em>speed</em> intervals such that the mileage of the vehicle described is 20 miles per gallon or less are: v ∈ [10 mi/h, 20 mi/h] ∪ [50 mi/h, 75 mi/h]
<h3>How to determine the range of speed associate to desired gas mileages</h3>
In this question we have a <em>quadratic</em> function of the <em>gas</em> mileage (g), in miles per gallon, in terms of the <em>vehicle</em> speed (v), in miles per hour. Based on the information given in the statement we must solve for v the following <em>quadratic</em> function:
g = 10 + 0.7 · v - 0.01 · v² (1)
An effective approach consists in using a <em>graphing</em> tool, in which a <em>horizontal</em> line (g = 20) is applied on the <em>maximum desired</em> mileage such that we can determine the <em>speed</em> intervals. The <em>speed</em> intervals such that the mileage of the vehicle is 20 miles per gallon or less are: v ∈ [10 mi/h, 20 mi/h] ∪ [50 mi/h, 75 mi/h].
To learn more on quadratic functions: brainly.com/question/5975436
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Answer:
Given A triangle ABC in which
∠C =90°,∠A=20° and CD ⊥ AB.
In Δ ABC
⇒∠A + ∠B +∠C=180° [ Angle sum property of triangle]
⇒20° + ∠B + 90°=180°
⇒∠B+110° =180°
∠B =180° -110°
∠B = 70°
In Δ B DC
∠BDC =90°,∠B =70°,∠BC D=?
∠BDC +,∠B+∠BC D=180°[ angle sum property of triangle]
90° + 70°+∠BC D =180°
∠BC D=180°- 160°
∠BC D = 20°
In Δ AC D
∠A=20°, ∠ADC=90°,∠AC D=?
∠A + ∠ADC +∠AC D=180° [angle sum property of triangle]
20°+90°+∠AC D=180°
110° +∠AC D=180°
∠AC D=180°-110°
∠AC D=70°
So solution are, ∠AC D=70°,∠ BC D=20°,∠DB C=70°
The angle between two vectors is:
CosФ = u - v / Magnitude(u) x magnitude(v)
Magnitude of u = SQRT(7^2 + -2^2) = SQRT(49 +4) = SQRT(53)
Magnitude of v = SQRT(-1^2 +2^2) = SQRT(1 +4) = SQRT(5)
u x v = (7 x -1) + (-2 x 2) = -7 + -4 = -11
cosФ = -11 / sqrt(53) x sqrt(5)
cosФ = -11sqrt265) / 265
Ф =cos^-1(-11sqrt265) / 265)
Ф=132.51 degrees.
Answer:
x = 18
Step-by-step explanation:
The parts of the triangles are proportional, so you can write ...
2x/80 = 27/60
Multiply by 120:
3x = 54
Divide by 3:
x = 18