Prove the following theorem indirectly. We will give you a start. Prove that a triangle cannot have two right angles. A triangle cannot have two right angles. Suppose a triangle had two right angles.

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Answer 1

Answer:

A triangle that has no two right angles is called a scalene triangle. If the triangle has two right angles, it is called isosceles triangle. A triangle with equal sides is called an equilateral triangle.

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Need a step by step on how to solve this please
Pls pls u guys I promised u will get 15 points if u answer this

C = 0.765x + 0.06(0.765x)

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Hey there!

$c=0.765x+0.06(0.765x)$

Add:

$c=0.765x+0.06(0.765x)\nc=0.8109x$

Hope this helps!

$AT2309$

PLS HELP FIRST CORRECT ANSWER GETS BRAINLIEST !! A pile of tailings from a gold dredge is in the shape of a cone. The diameter of the base is 34 feet and the height is 16 feet. Approximately, how many cubic feet of gravel is in the pile? Use π = 3.14.A. 14,527 ft³
B. 285 ft³
C. 4,840 ft³
D. 6,032 ft³

Answers

The volume of the cone is 4840 cubic ft if the diameter of the base is 34 feet and the height is 16 feet option (C) is correct.

What is a cone?

It is defined as a three-dimensional shape in which the base is a circular shape and the diameter of the circle decreases as we move from the circular base to the vertex.

$\rm V=\pi r^2(h)/(3)$

Volume can be defined as a three-dimensional space enclosed by an object or thing.

It is given that:

A pile of tailings from a gold dredge is in the shape of a cone.

The diameter of the base is 34 feet and the height is 16 feet.

As we know,

The volume of the cone is given by:

$\rm V=\pi r^2(h)/(3)$

r = 34/2 = 17 ft

h = 16 feet

Plug the above values in the formula:

$\rm V=\pi (17)^2(16)/(3)$

After solving:

V = 1541.33π cubic feet

Take π = 3.14

V = 1541.33(3.14) cubic feet

V = 4839.78 ≈ 4840 cubic ft

Thus, the volume of the cone is 4840 cubic ft if the diameter of the base is 34 feet and the height is 16 feet option (C) is correct.

Learn more about the cone here:

brainly.com/question/16394302

#SPJ5

Answer:

C: $V=4840 (2 s.f.)$

Step-by-step explanation:

The formula for the volume of a cone is:

$V= (1)/(3) \pi r^2h$

Therefore,

$V=(1)/(3)* 3.14*((34)/(2))^2* 16\n\nV=4840 (2 s.f.)$

Can someone pleeeeeasssse help with 3x-2/x^2-2x-3 divided by 1/x-3

Answers

If im correct your final answer should be 2x-3/(x+1)(x-3). #TeamAlvaxic

What is 90 out of 95 written as a fraction?

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$90\ out\ of\ 95 = (90)/(95)\ which\ simplifies\ to\ (18)/(19)$

90/95 or it can also be written as 90 over 95

You drive 6.0 km at 50 km/h and then another 6.0 km at 90 km/h. Your average speed over the 12 km drive will be _________

Answers

Answer:

$64.29\ (km)/(h)$

Step-by-step explanation:

hello

the speed is the relation between a distance traveled and the time used for it and it is given by the equation

$v =(displacement)/(time)\n (1)v=(d)/(t)$

Now, the average speed will be

$v_(average) =(displacement\ total)/(total\ time)\n (1)v=(d)/(t) \n(2)v_(average)=(d_(t) )/(t_(t) )$

Step 1

I drive 6.0 km at 50 km/h

from (1) it is possible to isolate t

$(1)v=(d)/(t)\nt=(d)/(v)$

Let

$v=50 (km)/(h) \nd=6.0 km\nt=(d)/(v) \nt=(6.0 km)/(50 (km)/(h))\nt_(1)=0.12\ hours\n$

Step 2

then I drive 6.0 km at 90 km/h

$v=90 (km)/(h) \nd=6.0 km\nt=(d)/(v) \nt=(6.0 km)/(90 (km)/(h))\nt_(2)=0.067\ hours\n$

Step 3

FInd total time and total displacement

$Total\ time=t_(1) +t_(2) \nt_(t) =0.12\ hours +0.067\ hours =0.187\ hours\n\ntotal\ distance\ =d_(1) +d_(2)=6.0\ km +6.0\ km= 12\ km\n$

step 4

finally, put these values into the equation(2)

$v_(average)=(d_(t) )/(t_(t) )\nv_(average)=(12\ km )/(0.187\ h )\nv_(average)=64.29\ (km)/(h)$

V=64.29 km/h

Have a good day

Answer:

64.2857 km/h

Step-by-step explanation:

Time taken to drive the first 6.0 km in hours = 6/50

Time take to drive another 6.0 km in hours = 6/90

Total time taken to drive 12 km in hours = (6/50) + (6/90)

= 6(50+90)/50x90

= 6x140/50x90

= 840/4500

The average speed over the 12 km is therefore 12 divided by 840/4500

= (12 x4500)/840

= 64.2857 km/h

During the "free banking" era, A. only the U.S. Treasury could issue currency. B. value and denominations were standard across the nation. C. the number of state chartered banks stayed the same. D. all banks were able to issue currency.

Answers

the answer is A. only the U.S. treasury could issue currency

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