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2 years ago

When √72 is expressed in simplest a√b form,

Mathematics

2 answers:

____ [38]2 years ago

7 0

Simplification :

$\sqrt{72}$

$\sqrt{2 \times 2 \times 2 \times 3 \times 3}$

$6 \sqrt{2}$

now, by equating the simplified term with a√b form we get :

$a = 6$

correct option - 1.) 6

weeeeeb [17]2 years ago

4 0

Answer:

Step-by-step explanation:

6 is the answer as the outcoming value is 6√2

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Rename 5/6 and 14/15 using the least common denominator.

Vedmedyk [2.9K]

Answer:

25/30 and 28/30

Step-by-step explanation:

1. To find the least common denominator, you need to find the least common multiple of the two denominators which are 6 and 15. Prime factorization of these numbers gives:

6 = 2 x 3

15 = 5 x 3

A number that would evenly divide both 6 and 15 must contain 2 x 3 x 5 which is 30.

Thus 30 is the least common denominator.

2. Now we need to somehow change both fractions to make them have a denominator of 30. To do this we can multiply by fractions with same numerator and denominator since that would be like multiplying by 1.

So, 5/6 x 5/5 = 25/30

And 14/15 x 2/2 = 28/30

8 0

3 years ago

Need answers for 70-75

adoni [48]

Answer:

70. 0

71. -54

72. 12

73. 86

74. 59

Step-by-step explanation:

To evaluate an expression, substitute specific values for the variables and simplify using Order of Operations.

70. c-3d becomes

$(1)-3(\frac{1}{3}) = 1-(\frac{3}{3}) = 1-1 = 0$

71. $x+x^3-4x^4$ becomes

$2+2^3-4(2)^4 = 2+8-4(16) = 2+8-64=10-64=-54$

72. $5a+3a^2$ becomes

$5(-3)+3(-3)^2 = -15 +3(9) = -15 + 27 = 12$

73. $F=\frac{9}{5}C+32$ becomes

$F=\frac{9}{5}(30)+32=\frac{270}{5}+32 = 54+32 = 86$

74. $F=\frac{9}{5}C+32$ becomes

$F=\frac{9}{5}(15)+32=\frac{135}{5}+32 = 27+32 = 59$

8 0

2 years ago

Lcm of 2²×3×5²and2²×3²×5

Bingel [31]

Answer:

580

Step-by-step explanation:

because you square the area and identify

4 0

3 years ago

What is the domain and range?

ycow [4]

Answer:

Domain- the set of possible values of the independent variable or variables of a function.

Range- the set of values that a given function can take as its argument varies.

8 0

3 years ago

Polygon ABCD is plotted on a coordinate plane and then rotated 90° clockwise about point C to form polygon A′B′C′D′. Match each

horrorfan [7]

The vertices of ABCD after 90 degrees clockwise rotation about point C are: A' (0,6), B' (3,7), C' (4,6) and D' (4,3)

<h3>How to match the vertices of the polygon?</h3>

The image of the polygon ABCD is not given; however, the question can still be answered because the coordinates are known.

The vertices of polygon ABCD are given as:

A = (4, 6)

B = (5, 3)

C = (4, 2)

D = (1, 2)

The rule of rotation about point C is:

(x,y) = (a + b - y, x + b - a)

Where:

(a, b) = (4, 2) --- the point of rotation.

So, we have:

(x,y) = (4 + 2 - y, x + 4 - 2)

(x,y) = (6 - y, x + 2)

The above means that:

A' = (6 - 6, 4 + 2) = (0,6)

B' = (6 - 3, 5 + 2) = (3,7)

C' = (6 - 2, 4 + 2) = (4,6)

D' = (6 - 2, 1 + 2) = (4,3)

Hence, the image of the rotation and their vertices (i.e. coordinates) are:

A' = (0,6)

B' = (3,7)

C' = (4,6)

D' = (4,3)