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3 years ago

8

A runner crossed the finish line at the time shown on the clock. The runners time was 2 hours and 45 minutes. What time did the

race begin?

1 answer:

dsp733 years ago

7 0

Answer:

The race began at 12.

Step-by-step explanation:

He finish at 2 hours and 45 minutes. If you go back 2 hours and 45 minutes from the clock it is 12.

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3 1/2 + (-7) x (2 2/3 + 1 1/2) = ?

mixer [17]

Answer: 77/3 or 25 2/3

Step-by-step explanation:

1. Convert mixed numbers into improper fractions:

3 1/2 = 3*2+1/2 = 7/2

2 2/3 = 2*3+2/3 = 8/3

1 1/2 = 1*2+1 = 3/2

2. Calculate within parentheses

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2 years ago

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olganol [36]

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What is the equation of the line that passes through the point (- 2, - 1) and has a slope of 5/2

Phoenix [80]

Answer:

$equation \: of \: a \: line \\ y - y1 = m(x - x1) \\ y - ( - 1) = \frac{5}{2}(x - ( - 2)) \\ 2(y + 1 )= 5(x + 2) \\ 2y + 2 = 5x + 10 \\ 2y - 5x - 8 = 0$

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2 years ago

PLEASE HELP!! <br> I need help on problems #5, #6, #9, #10<br> Picture attached <br> Thank you!

Romashka-Z-Leto [24]

Answer:

5. TS ≅ ML

6. ∠N≅∠U

9. m∠P = 80°

10. QR = 3

Step-by-step explanation:

The congruent symbol ≅ indicates that the two shapes have the same angle and side lengths.

5. TS are the third and second letter in RSTUV. The third and second letters in KLMNO are ML. TS ≅ ML

6. N is the fourth letter in KLMNO. The fourth letter is RSTUV is U. ∠N≅∠U

9. ∠P is congruent to ∠W. Notice they are both on the longer side and the wider angle. Since m∠P ≅ m∠W, and m∠W = 80°, then m∠P = 80°

10. QR is congruent to XY. They are both the shortest sides of the shape. QR ≅ XY and XY = 3, then QR = 3

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3 years ago

A flagpole broke in a storm. It was originally 81 8181 feet tall. 28 2828 feet are still sticking straight out of the ground, wh

solong [7]

Answer:

The end of the flagpole is 50.79 ft away from the base of the pole.

Step-by-step explanation:

The problem is represented by the diagram below.

The broken flagpole forms the shape of a right angled triangle. We need to find one of the sides of the triangle, the adjacent (x).

The hypotenuse is the broken part of the flagpole (53 ft), while the opposite is the part of the flagpole that is still stuck to the ground (28 ft).

Using Pythagoras theorem, we have that:

$hyp^2 = adj^2 + opp^2$

=> $53^2 = x^2 + 28^2$

$3364 = x^2 + 784\\\\=> x^2 = 3364 - 784\\\\x^2 = 2580\\\\x = \sqrt{2580}\\ \\x = 50.79 ft$

The end of the flagpole is 50.79 ft away from the base of the pole.

7 0

3 years ago