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3 years ago

If 1kg is the weight of seven apples, how many apples are in 2K grams?

Mathematics

2 answers:

pantera1 [17]3 years ago

6 0

1 kilogram contains = 7 apples

2 kilogram will contain = 7 *2 = 14 apples

So answer is 14 apples.

we just multiply the number of apples contained in 1 kilogram with the amount of kilogram we need to find and we get our answer.

14 apples

Elena L [17]3 years ago

5 0

Hey here is your answer

==========================

1kg = 1000g

given,

1kg = 7 apples

1000g = 7 apples

then,

2000g = 2 × 1000g

= 2 × 7 apples

= 14 apples

===========================

hope it may helpful to you

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Answer: 1/17

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3 years ago

Drake needs to drill holes in a wooden plank. The holes will have a diameter of 0.5 millimeters, with an error allowance of 0.05

Tanzania [10]

Answer:

|x − 0.5| = 0.05

Step-by-step explanation:

Here x represents the diameter,

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3 years ago

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Step-by-step explanation:

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3 years ago

A sapling grew 8 1/2 inches one year 10 3/4 inches the second year and 9 3/4 inches the third year how much did it grow during t

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3 years ago

Consider the points A(5, 3t+2, 2), B(1, 3t, 2), and C(1, 4t, 3). Find the angle ∠ABC given that the dot product of the vectors B

Vilka [71]

Answer:

66.42°

Step-by-step explanation:

<u>Given:</u>

A(5, 3t+2, 2)

B(1, 3t, 2)

C(1, 4t, 3)

BA • BC = 4

Step 1: Find t.

First we have to find vectors BA and BC. We do that by subtracting the coordinates of the initial point from the coordinates of the terminal point.

In vector BA B is the initial point and A is the terminal point.

BA = OA - OB = (5-1, 3t+2-3t, 2-2) = (4, 2, 0)

BC = OC - OB = (1-1, 4t-3t, 3-2) = (0, t, 1)

Now we can find t because we know that BA • BC = 4

BA • BC = 4

To find dot product we calculate the sum of the produts of the corresponding components.

BA • BC = (4)(0) + (2)(t) + (0)(1)

4 = (4)(0) + (2)(t) + (0)(1)

4 = 0 + 2t + 0

4 = 2t

2 = t

t = 2

Now we know that:

BA = (4, 2, 0)

BC = (0, 2, 1)

Step 2: Find the angle ∠ABC.

Dot product: a • b = |a| |b| cos(angle)

BA • BC = 4

|BA| |BC| cos(angle) = 4

To get magnitudes we square each compoment of the vector and sum them together. Then square root.

$|BA| = \sqrt{4^2 + 2^2 + 0^2} = \sqrt{20} = 2\sqrt{5}$

$|BC| = \sqrt{0^2 + 2^2 + 1^2} = \sqrt{5}$

$2\sqrt{5}\sqrt{5}\cos{(m\angle{ABC})} = 4$

$10\cos{(m\angle{ABC})} = 4$

$\cos(m\angle{ABC}) = \frac{4}{10}=\frac{2}{5}$

$m\angle{ABC} = cos^{-1}{(\frac{2}{5})}$

$m\angle{ABC} = 66.4218^{\circ}$

Rounded to two decimal places:

$m\angle{ABC} = 66.42^\circ$

3 0

2 years ago