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4 years ago

Which equation best describes the relationship between the number of students and the number of tables

Mathematics

1 answer:

Alenkasestr [34]4 years ago

4 0

Answer:

equation d

Step-by-step explanation:

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A rhombus ABCD has AB = 10 and m∠A = 60°. Find the lengths of the diagonals of ABCD.

melisa1 [442]

Three important properties of the diagonals of a rhombus that we need for this problem are:
1. the diagonals of a rhombus bisect each other
2. the diagonals form two perpendicular lines
3. the diagonals bisect the angles of the rhombus

First, we can let O be the point where the two diagonals intersect (as shown in the attached image). Using the properties listed above, we can conclude that ∠AOB is equal to 90° and ∠BAO = 60/2 = 30°.

Since a triangle's interior angles have a sum of 180°, then we have ∠ABO = 180 - 90 - 30 = 60°. This shows that the ΔAOB is a 30-60-90 triangle.

For a 30-60-90 triangle, the ratio of the sides facing the corresponding anges is 1:√3:2. So, since we know that AB = 10, we can compute for the rest of the sides.

$\overline{OB}:\overline{AB} = 1:2$
$\overline {OB}:10 = 1:2$
$\overline{OB} = \frac{1}{2}(10) = 5$

Similarly, we have

$\overline{AO}:\overline{AB} = \sqrt{3}:2$
$\overline {AO}:10 = \sqrt{3}:2$
$\overline{AO} = \frac{\sqrt{3}}{2}(10) = 5\sqrt{3}$

Now, to find the lengths of the diagonals,

$\overline{AD} = 2(\overline{AO}) = 10\sqrt{3}$
$\overline{BC} = 2(\overline{OB}) = 10$

So, the lengths of the diagonals are 10 and 10√3.

Answer: 10 and 10√3 units

8 0

3 years ago

I and m are parallel to each other

Misha Larkins [42]

And what I need more information to answer your question.

4 0

3 years ago

Read 2 more answers

Math Question:<br> Pls help I will do anything

kow [346]

Answer:

The probability of spinning red is $\frac{1}{3}$ , the probability of spinning a 1 is $\frac{1}{6}$ , the probability of spinning an odd number is $\frac{1}{2}$ , and the probability of spinning a 9 is 0.

Step-by-step explanation:

Since there are a total of 6 possible outcomes, you can use this in the following questions. There are 2 red tiles, so the probability of spinning red is $\frac{2}{6}$ or $\frac{1}{3}$ . Since there are only one 1 tile, the probability of spinning a 1 is $\frac{1}{6}$ . Since there are three odd numbers, the probability of spinning an odd number is $\frac{1}{2}$ . Since there are no 9 in the spinner, the probability of spinning a 9 is 0 or never.

5 0

3 years ago

Find the circumference and area of a circle with a diameter of 22 inches.

Katyanochek1 [597]

Answer:

$Circumference=22\pi\;inches \\\\Area =121\pi \;inches^2$

Step-by-step explanation:

Diameter of the circle= $22\;inches$

$Radius=\dfrac{22}{2} =11\;inches$

Circumference of a circle:

$=2\times \pi \times r$

$=2\times \pi \times 11\\\\=22\pi \;inches$

Area of the circle:

$=\pi \times r^2$

$=\pi \times 11\times 11\\\\=121\times \pi \\\\=121\pi \;inches^2$

8 0

3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago