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2 years ago

12

Please answer this thank you

1 answer:

exis [7]2 years ago

7 0

Step-by-step explanation:

option 1

option 3

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*****50 POINTSSSS*****

defon

Answer:

<h3>Given</h3>

m∠REG = 78°
mAR = 46°
ER ≅ GA

<h3>Solution</h3>

m∠GAR = 180° - m∠REG = 180° - 78° = 102° (supplementary angles sum to 180°)
m∠TAR = 1/2mAR = 1/2(46°) = 23° (tangent chord angle is half the size of intercepted arc)
m∠GAN = 180° - (m∠TAR + m∠GAR) = 180° - (23° + 102°) = 55° (straight angle is 180°)
mAG = 2m∠GAN = 2(55°) = 110°
mRE = mAG = 110° (as ER ≅ GA)
mGE = 360° - (mAG + mAR + mRE) = 360° - (110° + 46° + 110°) = 94° (full circle is 360°)

6 0

2 years ago

A pole that is 3.4 m tall casts a shadow that is 1.54 m long. At the same time, a nearby building casts a shadow that is 35.75 m

inna [77]

Answer:

The building would be 78.92 meters tall.

Step-by-step explanation:

If the shodow of a 3.4m tall pole is 1.54m tall, and you have the shadow's height, then you make a cross, multiply-divide ratio table.

5 0

3 years ago

A bucket that weighs 6 lb and a rope of negligible weight are used to draw water from a well that is 80 ft deep. The bucket is f

zaharov [31]

Answer:

3200 ft-lb

Step-by-step explanation:

To answer this question, we need to find the force applied by the rope on the bucket at time $t$

At $t=0, the weight of the bucket is 6+36=42 \mathrm{lb}$

After $t$ seconds, the weight of the bucket is $42-0.15 t \mathrm{lb}$

Since the acceleration of the bucket is the force on the bucket by the rope is equal to the weight of the bucket.

If the upward direction is positive, the displacement after $t$ seconds is $x=1.5 t$

Since the well is 80 ft deep, the time to pull out the bucket is $\frac{80}{2}=40 \mathrm{~s}$

We are now ready to calculate the work done by the rope on the bucket.

Since the displacement and the force are in the same direction, we can write

$W=\int_{t=0}^{t=36} F d x$

Use $x=1.5 t$ and $F=42-0.15 t$

$W=\int_{0}^{36}(42-0.15 t)(1.5 d t)$

$=\int_{0}^{36} 63-0.225 t d t$

$=63 \cdot 36-0.2 \cdot 36^{2}-0=3200 \mathrm{ft} \cdot \mathrm{lb}$

$=\left[63 t-0.2 t^{2}\right]_{0}^{36}$

$W=3200 \mathrm{ft} \cdot \mathrm{lb}$

4 0

3 years ago

Shown below is the solution to the linear program for finding Player A's optimal mixed strategy in a two-person, zero-sum game.

Luba_88 [7]

Answer:

Following are the answer to this question:

Step-by-step explanation:

For Option a:

Its optimal mixed approach for Player A's to Player A

A1 with a chance of .05 utilizing technique

Using the .60 chance strategy for A2

Use the .35 possibility strategy for A3

For Option b:

Optimal level mixed approach for team B:

Use the strategy for B1 with a probability of 50

Using the chance strategy for B2 at .50

no use strategy for B3

For Option c:

The estimated gain of Player A will be= 3.500

For Option d:

The estimated loss of Player B will be 3.500

8 0

3 years ago

How to Solve the problem 7/8 x 4 =

musickatia [10]

Answer:

The Answer is gonna be 7/8 x 4 =3.5

This is the right Answer :3

I hope you are having a great day ❤️❤️❤️

5 0

3 years ago

Read 2 more answers