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3 years ago

How do you approximate a square root to the nearest tenth?

Mathematics

1 answer:

adell [148]3 years ago

3 0

You would get the answer of the square root and round the answer to one decimal place, which is the tenth

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What must 111,111,102 be divided by to get 9 as the quotient and 0 as the remainder?

never [62]

ANSWER: 12,345,678

EXPLANATION:
we can work backwards by figuring out what 9(the answer) is multiplied by to get 111,111,102 (the starting number) we can do this by dividing by 9:

111,111,102/9 = 12,345,678

so: 9 * 12,345,678 = 111,111,102
which also means

111,111,102/12,345,678 = 9 (with 0 as a remainder)

5 0

3 years ago

Can someone check this, please (I do report as needed and I do mark brainliest)? Thank you!

vodomira [7]

Answer:

Step-by-step explanation:

cotx/cscx=cosx

Start on the left side.

cos(x) csc(x)

Apply the reciprocal identity to csc (x) .

cos(x) 1/sin(x)

Simplify

cos(x) 1/sin(x)cos(x)/sin(x)

Rewrite cos(x)/sin(x) as cot(x) .

cot(x)

Because the two sides have been shown to be equivalent, the equation is an identity.

cos(x) csc(x)=cot(x) is an identity

cot(x)−tan(x)/sin(x)cos(x)=csc^2(x)−sec^2(x) is an identity

You are correct :)

7 0

3 years ago

Find the product 7/10 x 3/6 (please help!) fractions

hichkok12 [17]

Answer:

7/10 x 3/6 = 21/60

21/60 in its simpliest form = 7/20

5 0

1 year ago

Read 2 more answers

3(x-1)=2x+9 how many solutions are there

Nostrana [21]

There are 4 solutions

5 0

3 years ago

Represent each of these relations on {1, 2, 3, 4} with a matrix (with the elements of this set listed in increasing order). a) {

Irina-Kira [14]

Answer:

for a a) {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}

$\left[\begin{array}{cccc}0&1&1&1\\0&0&1&1\\0&0&0&1\\0&0&0&0\end{array}\right]$

for b

b) {(1, 1), (1, 4), (2, 2), (3, 3), (4, 1)}

$\left[\begin{array}{cccc}1&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{array}\right]$

for c) {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

$\left[\begin{array}{cccc}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{array}\right]$

for d d) {(2, 4), (3, 1), (3, 2), (3, 4)}

$\left[\begin{array}{cccc}0&0&0&0\\0&0&0&1\\1&1&0&1\\0&0&0&0\end{array}\right]$

Step-by-step explanation:

in matrix, arrays are placed in rows , which represents the horizontal sides from left to right, while arrays in the column are placed vertically from top to bottom. Here, we placed the arrays in a 4x4 matrix

for a a) {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}

$\left[\begin{array}{cccc}0&1&1&1\\0&0&1&1\\0&0&0&1\\0&0&0&0\end{array}\right]$

for b

b) {(1, 1), (1, 4), (2, 2), (3, 3), (4, 1)}

$\left[\begin{array}{cccc}1&0&0&1\\0&1&0&0\\0&0&1&0\\1&0&0&0\end{array}\right]$

for c) {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

$\left[\begin{array}{cccc}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{array}\right]$

for d d) {(2, 4), (3, 1), (3, 2), (3, 4)}

$\left[\begin{array}{cccc}0&0&0&0\\0&0&0&1\\1&1&0&1\\0&0&0&0\end{array}\right]$

7 0

3 years ago