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2 years ago

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Converting from the US customary system to the metric system, 16 fluid ounces is approximately equal to a) 1 gallon. O b) 160 mi

lliliters. c) 480 milliliters. O d) 960 milliliters.

1 answer:

zhuklara [117]2 years ago

6 0

Answer:

1 oz = 29.574 ml

16 oz = About 480 ml, so, yes, the answer is C.

Step-by-step explanation:

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If rs || mn ,then the value of a is

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You need at least 30 cubic feet of sand to fill a sand box. Each box contains 2.5 cubic feet of sand. Write and solve an inequal

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3 years ago

The 2 main support rods for a barbeque grill are each 7.5 cm from the centre of the

sukhopar [10]

Answer:

58.1 cm

Step-by-step explanation:

The length of each support rod can be found using the Pythagorean theorem. The geometry can be modeled by a right triangle, such that the distance from centre is one leg and half the length of the rod is the other leg of a triangle with hypotenuse equal to the radius of the grill.

__

<h3>Pythagorean theorem</h3>

The theorem tells us that the sum of the squares of the legs of a right triangle is the square of the hypotenuse. For legs a, b and hypotenuse c, this is ...

c² = a² +b²

<h3>application</h3>

For the geometry of the grill, we can define a=7.5 and c=30. Then b will be half the length of the support rod.

30² = 7.5 +b²

b² = 900 -56.25 = 843.75

b = √843.75 ≈ 29.0473

The length of each support rod is twice this value, so ...

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Each support rod is about 58.1 cm long.

7 0

2 years ago

Find the derivative of the function at P 0 in the direction of A. f(x,y,z) = 3 e^x cos(yz), P0 (0, 0, 0), A = - i + 2 j + 3k

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The derivative of $f(x,y,z)$ at a point $p_0=(x_0,y_0,z_0)$ in the direction of a vector $\vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k$ is

$\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}$

We have

$f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k$

and

$\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}$

Then the derivative at $p_0$ in the direction of $\vec a$ is

$3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}$

3 0

4 years ago