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3 years ago

What is the perimeter of a square

Mathematics

2 answers:

bearhunter [10]3 years ago

8 0

Add up the lengths of all the sides or one side length multiplied by 4!

Lera25 [3.4K]3 years ago

6 0

The perimeter of a square is P=4a

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ASAP Help again please

Arisa [49]

Answer:

Step-by-step explanation:

c. 0.875

140,000,000x0.25=35,000,000

35,000,000/40,000,000=0.875

4 0

3 years ago

In the accompanying diagram, lines a and b are parallel, and lines c and d are transversals. 5 a 6 3 4 7/8 be Which angle is con

OLga [1]

Answer:

Step-by-step explanation:

7 0

3 years ago

Find the value of x which satisfies the following equation.<br> log2(x−1)+log2(x+5)=4

weqwewe [10]

$\quad \huge \quad \quad \boxed{ \tt \:Answer }$

$\qquad \tt \rightarrow \: x = 3$

____________________________________

$\large \tt Solution \: :$

$\qquad \tt \rightarrow \: log_{2}(x - 1) log_{2}(x + 5) = 4$

$\qquad \tt \rightarrow \: log_{2} \{(x - 1)(x + 5) \} = 4$

[ log (x) + log (y) = log (xy) ]

$\qquad \tt \rightarrow \: ( x - 1)(x + 5) = {2}^{4}$

$\qquad \tt \rightarrow \: {x}^{2} + 5x - x - 5 = 16$

$\qquad \tt \rightarrow \: {x}^{2} + 4x - 5 - 16 = 0$

$\qquad \tt \rightarrow \: {x}^{2} + 4x -21 = 0$

$\qquad \tt \rightarrow \: {x}^{2} + 7x - 3x - 21 = 0$

$\qquad \tt \rightarrow \: x(x + 7) - 3(x + 7) = 0$

$\qquad \tt \rightarrow \: (x + 7)(x - 3) = 0$

$\qquad \tt \rightarrow \: x = - 7 \: \: or \: \: x = 3$

The only possible value of x is 3, since we can't operate logarithm with a negative integer in it.

$\qquad \tt \rightarrow \: x = 3$

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

4 0

2 years ago

Determine whether the given vectors are orthogonal, parallel or neither. (a) u=[-3,9,6], v=[4,-12,-8,], (b) u=[1,-1,2] v=[2,-1,1

nevsk [136]

Answer:

a) $u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

b) $u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

Since the angle between the two vectors is not 0 or 180 degrees we can conclude that are either.

c) $u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

Step-by-step explanation:

For each case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

Part a

u=[-3,9,6], v=[4,-12,-8,]

The dot product on this case is:

$u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

And finally we can calculate the angle between the vectors like this:

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

Part b

u=[1,-1,2] v=[2,-1,1]

The dot product on this case is:

$u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

And finally we can calculate the angle between the vectors like this:

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

Since the angle between the two vectors is not 0 or 180 degrees we can conclude that are either.

Part c

u=[a,b,c] v=[-b,a,0]

The dot product on this case is:

$u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

5 0

3 years ago

Read 2 more answers

7. Which set of ordered pairs represents a function?

Likurg_2 [28]

Answer:

Step-by-step explanation:

to be a function, every x has to have exactly one y, so there can't be the same number x twice wirh different y's

5 0

3 years ago