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4 years ago

?>?>?>>??>Find the area of the kite.

Mathematics

1 answer:

sineoko [7]4 years ago

8 0

Check the picture below.

$\bf \stackrel{\textit{area of triangles on the left}}{2\left[\cfrac{1}{2}(2)(3) \right]}+\stackrel{\textit{area of triangles on the right}}{2\left[\cfrac{1}{2}(4)(3) \right]}\implies 6+12\implies 18$

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Consider the following set of equations, where s, s0, x and r have units of length, t has units of time, v has units of velocity

Anna11 [10]

Answer:

Step-by-step explanation:

For an equation to be dimensionally correct the dimension of quantities on both sides of equation must be same.

Also, two physically quantities can only be added or subtracted only when their dimension are same.

here all option are dimensionally correct except the 5th option where

dimension of t= [T] whereas dimension of a/v is $\frac{LT^{-2}}{LT^{-1}}$

= T^{-1}

since, the dimension of quantities on either sides of equation are not the same the equation is dimensionally is incorrect.

8 0

3 years ago

I need help.. i really want to go sleep.. thank you so much...

Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Now find $\sum\limits_{i=3}^{11}\frac{1}{i}$

Expanding the summation we get

$\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Comparing the two series we get,

$\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$ so the given equality is true.

2) Given $\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}$

Verify the above equality is true or false

Now find $\sum\limits_{k=0}^4\frac{3k+3}{k+6}$

Expanding the summation we get

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}$

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}$

now find $\sum\limits_{i=1}^3\frac{3i}{i+5}$

Expanding the summation we get

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}$

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}$

Comparing the series we get that the given equality is false.

ie, $\sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}$

6 0

4 years ago

How many times do you have to regroup to subtract 847 -268

navik [9.2K]

Answer:

= 3 R 43

= 3 43/268

847 divided by 268 equals

3 with a remainder of 43

5 0

3 years ago

What are the steps involved in the financial planning process?

rosijanka [135]

Answer:

determine current financial situation

develop your financial goals

identify alternative course of action

evaluate alternatives

Create and implement your financial action plan

review and revise the financial plan.

Hope this helps have a great day! btw I have a question plz help me its due tonight.

5 0

3 years ago

Find a value for x if 1/x > x

Ymorist [56]

Since x can't be zero (it sits at the denominator of the left hand side), let's split cases:

If x is positive

It this case, we can multiply both sides by x without changing the inequality sign:

$\dfrac{1}{x} > x \iff 1 > x^2 \iff x \in (-1,1)$

But since we're assuming that x is positive, we can only accept the numbers between 0 and 1.

If x is negative

It this case, in order to multiply both sides by x we have to change the inequality sign:

$\dfrac{1}{x} > x \iff 1 < x^2 \iff x \in (-\infty,-1) \cup (1,\infty)$

But since we're assuming that x is negative, we can only accept the numbers which are smaller than -1.

So, putting all the pieces together, we have that the solution is given by the following interval:

$(-\infty, -1) \cup (0,1)$

6 0

3 years ago