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3 years ago

8

BRAINIESTT ASAP! PLEASE HELP ME :)

1 answer:

andreev551 [17]3 years ago

5 0

All sides are defined, so we use the <em>Law</em><em> </em><em>of</em><em> </em><em>Cosines</em>. Plus, the angle opposite from side <em>a</em><em> </em>is the largest.

I am joyous to assist you anytime.

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ANSWER THIS URGENTLY

neonofarm [45]

I think B if not type 1/3 pi 6 squared and add 6.71 or something with the 6.71

8 0

2 years ago

The fence along the length of a ranch measures 4,450.5 meters and the fence along the width of a ranch measures 2,890 meters.

lawyer [7]

Given:Length = 4,450.5 metersWidth = 2,890 meters
1 kilometer = 1,000 meters
Length: 4,450.5 meters * 1km/1,000 meters = 4,450.5 km/1,000 = 4.4505km
Width: 2,890 meters * 1km/1,000 meters = 2,890 km/1,000 = 2.89 km
4.4505 km - 2.89 km = 1.5605 km
The length of the fence is 1.5605 km longer than the width of the fence.

3 0

3 years ago

Hurry 50 points for correct answer

Ad libitum [116K]

Answer: <em>"7 is a solution to the original equation. The value –1 is an extraneous solution."</em>

Step-by-step explanation:

The equation $x=\sqrt{6x+7}$ can be solved by squaring both sides:

$x^2 = 6x + 7\\\\x^2 - 6x - 7 = 0\\\\(x-7)(x+1) = 0$

We can see that -1 and 7 are solutions, but make sure they are not extraneous by substituting them in the original equation:

$1 = \sqrt{-1}\\ 7 = \sqrt{49}$

The square root of 49 equals 7, but the square root of -1 is an imaginary number.

The correct choice is <em>"7 is a solution to the original equation. The value –1 is an extraneous solution."</em>

4 0

3 years ago

Suppose a class of 50 students has 20 males and 30 females. The insturctor will pick 1 (different) student to compete in 3 diffe

Pie

Answer:

Step-by-step explanation:

Total = 50

Males = 20

Females = 30

a) P(exactly 1 female) = P(1 female , 2 male)

(30C1 * 20C2) / (50C3) = 30*190/19,600 = 0.2908

b) P(at least 2 males) = 1 - (P(0 males) + P(1 males))

= 1 - 30C3 * 20C0/50C3 - 30C2 * 20C1/50C3

= 1 - 4060*1/19,600 - 435*20/19,600 = 6840/19600 = 0.3489

4 0

3 years ago

Read 2 more answers

At 2:00pm a car's speedometer reads 20mph, and at 2:10pm it reads 30mph.

Dmitry [639]

Over this 10-minute interval, the car's average acceleration is

$\dfrac{30\,\mathrm{mph}-20\,\mathrm{mph}}{10\,\mathrm{min}}=\dfrac{10\,\mathrm{mph}}{\frac16\,\mathrm h}=60\dfrac{\rm mi}{\mathrm h^2}$

The MVT says that at some point during this 10-minute interval, the car must have had an acceleration of 60 mi/h^2.

6 0

3 years ago