Answer:
D
Step-by-step explanation:
If you figure out the median you'll see that for 8th grafe its 21 and 7th 25 therefore 4 points higher
When two parallel lines are intersected by a transversal, the same-side exterior angles are supplementary. That means that their sum is 180.
Using that logic, if the two roads were parallel, then the sum of their same-side exterior angles will add up to 180. Yet their same-side exterior angles add up to 170 (130 + 40 = 170), hence they can't be parallel.
See the drawing attached below.
Using supplmenatry angles (two angles whose sum of measures add up to 180 or a straight line), we can say that:
m<DIE + m<HID = 18
40 + m<HID = 180
m<HID = 140
Similarly:
m<BHC + m<CHI = 180
130 + m<CHI = 180
m<CHI = 50
Using verticle angles therome, (when two lines intersect, the angles opposite to eachother are congruent, or have the same measure), we can say that:
m<DIE = m<GIH = 40
m<GIE = m<HID = 140
m<CHI = m<AHB = 50
m<BHC = m<AHI = 130
Answer:
He drove the rest of the trip at 30 miles per hour
Step-by-step explanation:
The rule of the distance is D = v × t, where
∵ It is 170 miles from Bruce's house to the city where his brother lives
∴ D = 170 miles
∵ He drove for 2 hours at 55 miles per hour
∴ t1 = 2 hours and v1 = 55 miles/hour
→ By using the rule above find the distance of this part of his trip
∴ D1 = 55 × 2 = 110 miles
∵ The rest of the trip took 2 hours
∴ t2 = 2 hours
∵ D = D1 + D2
→ Substitute the values of D and D1 to find D2
∴ 170 = 110 + D2
→ Subtract 110 from both sides to find D2
∴ 60 = D2
∴ D2 = 170 - 110 = 60 miles
∵ The rest of the trip is 60 miles
∵ It took 2 hours
∵ D2 = v2 × t2
∴ 60 = v2 × 2
→ Divide both sides by 2
∴ 30 = v2
∴ v2 = 30 miles/hour
∴ He drove the rest of the trip at 30 miles per hour
Answer:
200 cm²
Step-by-step explanation:
Find the areas of all of the faces
17 × 2 = 34 (Top Face)
8 × 15 = 120 (Both Triangles)
15 × 3 = 30 (Bottom Face)
8 × 2 = 16 (Back Face)
Add them all together
120 + 34 + 30 + 16 = 200 cm²

has gradient

which at the point (-1, 4, 3) has a value of

I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say
, in which case we have

Then the derivative of
at (-1, 4, 3) in the direction of
is
