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horrorfan [7]
3 years ago
9

Round to the place value of the underline digit 4 in the number 641,267

Mathematics
1 answer:
Rom4ik [11]3 years ago
4 0

Note the place value of 4. Look at the place value next to the 4. It is a 1. Because 1 is less than 5, round down.

641267 rounded to the nearest ten thousands (as 4 is in the ten thousands place value), is 640000.

hope this helps

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Parallelogram ABCD has vertices at A(2,-3), B(8,-6), Cl14,-3), and D(8,0)
OlgaM077 [116]

Answer:

The correct option is C.

Step-by-step explanation:

The vertices of ABCD are A(2,-3), B(8,-6), C(14,-3), and D(8,0).

Distance formula:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

The length of sides are

AB=\sqrt{(8-2)^2+(-6+3)^2}=\sqrt{36+9}=\sqrt{45}

BC=\sqrt{(14-8)^2+(-3+6)^2}=\sqrt{36+9}=\sqrt{45}

CD=\sqrt{(8-14)^2+(0-3)^2}=\sqrt{36+9}=\sqrt{45}

AD=\sqrt{(8-2)^2+(0+3)^2}=\sqrt{36+9}=\sqrt{45}

Since length of all sides are equation therefore the given parallelogram cannot be a rectangle.

Slope formula:

m=\frac{y_2-y_1}{x_2-x_1}

Slope of AB is

m_1=\frac{-6-(-3)}{8-2}=\frac{-3}{6}=\frac{-1}{2}

Slope of BC is

m_2=\frac{-3-(-6)}{14-8}=\frac{3}{6}=\frac{1}{2}

Since the slopes of two consecutive sides are not opposite reciprocals, therefore the given parallelogram is a rhombus. Option C is correct.

8 0
3 years ago
Find the volume of a rectangular prism that is 10 cm long, 3 cm wide, and 2 cm high. Please draw and label the prism​
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Answer:

Volume = Length × Width × Height

Volume = 10 × 3 × 2

Volume = 60 cm

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A bag contains 6 blue marbles, 4 red marbles, and 2 yellow marbles. What is the probability of selecting a blue marble, replacin
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Minimizing Packaging Costs If an open box has a square base and a volume of 107 in.3 and is constructed from a tin sheet, find t
uysha [10]

Answer:

The dimensions of the box that minimizes the amount of material of construction is

Square base = (5.98 × 5.98) in²

Height of the box = 2.99 in.

Step-by-step explanation:

Let the length, breadth and height of the box be x, z and y respectively.

Volume of the box = xyz = 107 in³

The box has a square base and an open top.

x = z

V = x²y = 107 in³

The task is to minimize the amount of material used in its construction, that is, minimize the surface area of the box.

Surface area of the box (open at the top) = xz + 2xy + 2yz

But x = z

S = x² + 2xy + 2xy = x² + 4xy

We're to minimize this function subject to the constraint that

x²y = 107

The constraint can be rewritten as

x²y - 107 = constraint

Using Lagrange multiplier, we then write the equation in Lagrange form

Lagrange function = Function - λ(constraint)

where λ = Lagrange factor, which can be a function of x and y

L(x,y,z) = x² + 4xy - λ(x²y - 107)

We then take the partial derivatives of the Lagrange function with respect to x, y and λ. Because these are turning points, at the turning points each of the partial derivatives is equal to 0.

(∂L/∂x) = 2x + 4y - 2λxy = 0

λ = (2x + 4y)/2xy = (1/y) + (2/x)

(∂L/∂y) = 4x - λx² = 0

λ = (4x)/x² = (4/x)

(∂L/∂λ) = x²y - 107 = 0

We can then equate the values of λ from the first 2 partial derivatives and solve for the values of x and y

(1/y) + (2/x) = (4/x)

(1/y) = (2/x)

x = 2y

Hence, at the point where the box has minimal area,

x = 2y

Putting these into the constraint equation or the solution of the third partial derivative,

x²y - 107 = 0

(2y)²y = 107

4y³ = 107

y³ = (107/4) = 26.75

y = ∛(26.75) = 2.99 in.

x = 2y = 2 × 2.99 = 5.98 in.

Hence, the dimensions of the box that minimizes the amount of material of construction is

Square base = (5.98 × 5.98) in²

Height of the box = 2.99 in.

Hope this Helps!!!

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