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3 years ago

1658.013 to the nearest hundred

Mathematics

2 answers:

tiny-mole [99]3 years ago

4 0

The answer would be 1700.000

laila [671]3 years ago

3 0

How to round to nearest hundred:

16xx.xxx

If xx.xxx rounds up (above 50), in which it does, It raises from 16xx to 1700.

Therefore the answer is 1700.000

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What is two thirds divided by seven

Westkost [7]

Answer:

$\frac{2}{21}$ i think

Step-by-step explanation:

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2 years ago

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Haley makes earrings and packages them into cube boxes that measure 1/6 foot wide. how many 1/6 foot cubic boxes can she fit int

navik [9.2K]

Answer: $28\ cubes$

Step-by-step explanation:

The volume of a cube can be found with this formula:

$V_{(c)}=s^3$

Where "s" is the lenght of any edge of the cube.

The formula for calculate the volume of a rectangular prism is:

$V_{rp}=lwh$

Where "l" is the lenght, "w" is the width and "h" is the height.

We need to find the volume of a cube box:

$V_1=s^3=(\frac{1}{6}ft)^3=\frac{1}{216}ft^3$

To find the volume of the shipping box, first we must convert the mixed number to an improper fraction:

$1\frac{1}{6}=\frac{(6*1)+1}{6}=\frac{7}{6}$

Then the volume of the shipping box is:

$V_2=lwh\\\\V_2=(\frac{7}{6}ft)(\frac{1}{3}ft)(\frac{1}{3}ft)=\frac{7}{54}ft^3$

Now, in order to find the number of cube boxes can Haley fits into a shipping box, you must divide the the volume of the shipping box by the volume of one cube. This is:

$\frac{\frac{7}{54}ft^3}{\frac{1}{216}ft^3}=28$

6 0

3 years ago

!!!! HELP PLEASE !!!! WILL GIVE BRAINLIEST!!!!!!!!!

AnnyKZ [126]

Answer:

Step 1 is incorrect because it was mutiplied. Instead, you should divide.

Step-by-step explanation:

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2 years ago

Find the quotient of these Complex Numbers. <br><br><br> (5-i) / (3+2i)

JulsSmile [24]

Let the given complex number

z = x + ix = $\dfrac{5-i}{3+2i}$

We have to find the standard form of complex number.

Solution:

∴ x + iy = $\dfrac{5-i}{3+2i}$

Rationalising numerator part of complex number, we get

x + iy = $\dfrac{5-i}{3+2i}\times \dfrac{3-2i}{3-2i}$

⇒ x + iy = $\dfrac{(5-i)(3-2i)}{3^2-(2i)^2}$

Using the algebraic identity:

(a + b)(a - b) = $a^{2}$ - $b^{2}$

⇒ x + iy = $\dfrac{15-10i-3i+2i^2}{9-4i^2}$

⇒ x + iy = $\dfrac{15-13i+2(-1)}{9-4(-1)}$ [ ∵ $i^{2} =-1$ ]

⇒ x + iy = $\dfrac{15-2-13i}{9+4}$

⇒ x + iy = $\dfrac{13-13i}{13}$

⇒ x + iy = $\dfrac{13(1-i)}{13}$

⇒ x + iy = 1 - i

Thus, the given complex number in standard form as "1 - i".

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3 years ago

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zhuklara [117]

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3 years ago