$2.70 for 15 ounces please help asap

13 inches = __ ft<br> PLS HELP PLS

Radda [10]

Answer:

1.083

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Brad has two alloys containing copper and silver. Alloy A contains 50% pure silver, and alloy B contains 75% pure silver. He wan

oee [108]

Answer:

I think the answer would be

Alloy A 30%

Alloy B 40%

The reason I say that is because Alloy B has a higher amount of pure silver so yeah

I hope this helps mark me brainliest and tell me if I'm wrong please so Then I could fix it.

P.S. Wegnerkolmp2741o is gonna answer so I would go with his

8 0

3 years ago

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(PICTURE) Can someone please help me out Im reporting trolls. Marking brainliest

Goshia [24]

Answer:

smart dot company would be c=.50t+12

communication plus would be c=2.50t

Step-by-step explanation:

smart dot starts with $12 a month and adds $0.50 for each hour of internet use. communication plus has no starting fee so it just adds $2.50 for each hour of internet use.

So then for the table you would put the number given for time in for the letter t and find what c would be. This would then find your cost for that amount of hours. To get you started smart dot company starts off with 0 hours so put that in for t being c=0.50(0)+12 and c=12. Communication plus also starts with 0 so it would be c=2.50(0) and c=0

Then you graph once you have both values you can graph having x=time and y=cost

Let me know if this helps!

4 0

3 years ago

Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point

Maksim231197 [3]

Answer: $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

or (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

$\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]$

Taking square on both the sides , we get

$(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0$

Using quadratic formula : $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}$

so, $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

since $\sqrt{3}=1.732$

so, $(6+7(1.732),0)\text{ and }(6-7(1.732),0)$ are the required points.

i.e. (18.124,0) and ( -6.124,0) are the required points.

3 0

3 years ago

A North American river otter wants to cross the North Saskatchewan river. It can swim at a speed of 2.75 m/s. In that location,

insens350 [35]

Answer:

Step-by-step explanation:

time to cross will be the distance divided by the velocity in that direction.

a) t = d/v = 45/2.75 = 16.363636... 16.4 s

In the same time that the otter is swimming cross stream, the stream is also moving perpendicularly at its own pace.

b) d = vt = 2.5(16.36) = 40.9090... 40.9 m

As velocity is a vector, An observer on the bank would see the velocity as the vector addition of the cross stream and down stream velocities

c) v = √(2.75² + 2.5²) = 3.716517... 3.72 m/s magnitude

θ = arctan2.5/2.75 = 42.2736... 42.3° downstream of a line straight across

7 0

3 years ago