What is this question to the nearest mile ?

Mathematics

2 answers:

stiv31 [10]3 years ago

8 0

50 kilometers is 31.0686 miles, so 31.0696 to the nearest mile is 31 miles

Sedbober [7]3 years ago

4 0

The answer rounding up to the closest mile is 31 miles

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A brace for a shelf has the shape of a right triangle. Its hypotenuse is 10 inches long and the two legs are equal in length. Ho

vagabundo [1.1K]

This is a classic example of a 45-45-90 triangle: it's a right triangle (one angle of 90) & two other sides of the same length, which means two angles of the same length (and 45 is the only number that will work). With a 45-45-90 triangle, the lengths of the legs are easy to determine:
45-45-90
1-1-sqrt2
Where the hypotenuse corresponds to sqrt2.
Now, your hypotenuse is 10.
To figure out what each leg is, divide 10/sqrt2 (because sqrt2/sqrt2 = 1, which is a leg length in the explanation above).
Problem: you can't divide by radicals. So, we'll have to rationalize the denominator:
(10•sqrt2)/(sqrt2•sqrt2)
This can be rewritten:
10sqrt2/sqrt(2•2)
=10sqrt2/sqrt4
=10sqrt2/2
=5sqrt2 Hope this helps!!

3 0

3 years ago

A home has three bedrooms, 2 baths, a pool, and an attached workshop. The asking price is $ 450,000; the square footage is 2,255

castortr0y [4]

Answer:

$ 199.56 per sq ft

Step-by-step explanation:

price / ft^2 = 450 000 / 2255 = 199.56

8 0

2 years ago

The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to

Firdavs [7]

Rewrite the boundary lines y = -1 - x and y = x - 1 as functions of y :

y = -1 - x ==> x = -1 - y

y = x - 1 ==> x = 1 + y

So if we let x range between these two lines, we need to let y vary between the point where these lines intersect, and the line y = 1.

This means the area is given by the integral,

$\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy$

The integral with respect to x is trivial:

$\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy$

For the remaining integral, integrate term-by-term to get

$\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}$

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line y = 1 and (0, -1)), so its area is 1/2*4*2 = 4.

6 0

3 years ago

3z-4=6z-17 how do solve

Maru [420]

3z-4=6z-17
-3z-4=-17 (subtracted the 6z from both sides)
-3z=-13 (added the 4 to both sides)
z= $\frac{-13}{-3z}$ (divided both sides by -3z)
z= $\frac{13}{3}$