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tiny-mole [99]
4 years ago
15

What is the simplified version of the square root of 64/9?

Mathematics
2 answers:
ASHA 777 [7]4 years ago
7 0

\sqrt{ \frac{64}{9} } =  \frac{8}{3}
kramer4 years ago
3 0

You can distribute square roots between numerator and denominator:

\sqrt{\dfrac{64}{9}} = \dfrac{\sqrt{64}}{\sqrt{9}}

Now, the square root of a number y is a number x such that x^2 = y

So, \sqrt{64} is a number that, when squared, gives 64. This number is 8, since 8^2 = 64

For the same reason, \sqrt{9}=3

So, you have

\sqrt{\dfrac{64}{9}} = \dfrac{\sqrt{64}}{\sqrt{9}} = \dfrac{8}{3}

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8

Step-by-step explanation:

(15-4x)+5

(15-4(3))+5

(15-12)+5

3+5=8

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3 years ago
A new Community Center is being built in Oak Valley. The perimeter of the rectangular playing field is 382 yards. The length of
vesna_86 [32]

Answer:

The width is 50 yards and the length is 141 yards.

Step-by-step explanation:

Let's call: L the length of the field and W the width of the field.

From the sentence, the perimeter of the rectangular playing field is 382 yards we can formulate the following equation:

2L + 2W = 382

Because the perimeter of a rectangle is the sum of two times the length with two times the width.

Then, from the sentence, the length of the field is 9 yards less than triple the width, we can formulate the following equation:

L = 3W - 9

So, replacing this last equation on the first one and solving for W, we get:

2L + 2W = 382

2(3W - 9) + 2W = 382

6W -18 +2W = 382

8W - 18 = 382

8W = 382 + 18

8W = 400

W = 400/8

W = 50

Replacing W by 50 on the following equation, we get:

L = 3W - 9

L = 3(50) - 9

L = 141

So, the width of the rectangular field is 50 yards and the length is 141 yards.

3 0
3 years ago
Determine whether the points lie on straight line.
wel

Answer:

A) Non-collinear- does not lies on straight line

B) Collinear- lie on straight line

Step-by-step explanation:

We have to check collinearity of three points.

The points (x_1, y_1, z_1),(x_2, y_2, z_2),(x_3, y_3, z_3) are said to be collinear if,

\left[\begin{array}{ccc}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{array}\right] = 0

A) A(2, 5, 3), B(3, 7, 1), C(1, 4, 4)

\left[\begin{array}{ccc}2&5&3\\3&7&1\\1&4&4\end{array}\right] \\\\=2(28-4)-5(12-1)+3(12-7)\\= 8 \neq 0

Thus, the given points are not collinear.

B) D(0,-5,5), E(1,-2,4), F(3,4,2)

\left[\begin{array}{ccc}0&-5&5\\1&-2&4\\3&4&2\end{array}\right] \\\\=0(-4-16)+5(2-12)+5(4+6)\\=0

Thus, the given points are collinear.

6 0
3 years ago
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