All categories

2 years ago

What is the square root of 30,300

Mathematics

2 answers:

I am Lyosha [343]2 years ago

6 0

√30,300 = 174.06895185529210 or simply just 174.1 rounded. Therefore your answer to √of 30,300 is 174.1

garri49 [273]2 years ago

5 0

174.068951855thats the answer

You might be interested in

Matrix A has order 4 x 8, Matrix B has order 4 x 16.

Scorpion4ik [409]

Matrix A order is 4x8.
Matrix B order is 4x16.
AB cannot result because an (m x n) can only be multiplied by an (n x p) matrix and orders don't match the requirements.
BA has the same negative result.

3 0

3 years ago

Please answer- what is the length of BW? ANSWER DIRECTLY NO LINK ( will give brainlist to whoever answers)

Yuki888 [10]

Answer:

148.49

Step-by-step explanation:

105 x √2 = 148.49

have a great day

4 0

2 years ago

20 points and I will mark brainliest!!

JulijaS [17]

Answer:20.5

Step-by-step explanation:

4 0

3 years ago

Solve this please…..

Paraphin [41]

Answer:

17,12,15,14,12 answer 90

6 0

2 years ago

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation $(x+0,y-2)$ will map vertices of the square into points

$(2,1)\rightarrow (2,-1);$
$(1,2)\rightarrow (1,0);$
$(2,3)\rightarrow (2,1);$
$(3,2)\rightarrow (3,0).$

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

$(2,-1)\rightarrow (2,3);$
$(1,0)\rightarrow (1,2);$
$(2,1)\rightarrow (2,1);$
$(3,0)\rightarrow (3,2)$

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation $(x+2,y-0)$ will map vertices of the square into points

$(2,1)\rightarrow (4,1);$
$(1,2)\rightarrow (3,2);$
$(2,3)\rightarrow (4,3);$
$(3,2)\rightarrow (5,2).$

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

$(4,1)\rightarrow (2,1);$
$(3,2)\rightarrow (3,2);$
$(4,3)\rightarrow (2,3);$
$(5,2)\rightarrow (1,2)$

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation $(x+3,y+3)$ will map vertices of the square into points

$(2,1)\rightarrow (5,4);$
$(1,2)\rightarrow (4,5);$
$(2,3)\rightarrow (5,6);$
$(3,2)\rightarrow (6,5).$

2nd transformation = reflection over y = -x + 7 will map vertices into points

$(5,4)\rightarrow (3,2);$
$(4,5)\rightarrow (2,3);$
$(5,6)\rightarrow (1,2);$
$(6,5)\rightarrow (2,1)$

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation $(x-3,y-3)$ will map vertices of the square into points

$(2,1)\rightarrow (-1,-2);$
$(1,2)\rightarrow (-2,-1);$
$(2,3)\rightarrow (-1,0);$
$(3,2)\rightarrow (0,-1).$

2nd transformation = reflection over y = -x + 2 will map vertices into points

$(-1,-2)\rightarrow (4,3);$
$(-2,-1)\rightarrow (3,4);$
$(-1,0)\rightarrow (2,3);$
$(0,-1)\rightarrow (3,2)$

These points are not the vertices of the initial square.

5 0

3 years ago