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Marina CMI [18]
2 years ago
15

I need help with this pls

Mathematics
1 answer:
antoniya [11.8K]2 years ago
7 0
I'll explain it simply for you

1st question
Of course you know phythagoras theorm
You even wrote it up there
It states that the sum of the square of the two sides of an equilateral triangle is equal to the square of the hypotenuse
{a}^{2}  +  {b}^{2}   =  {c}^{2}
Where C is the hypotenuse
*NOTE* :
HYPOTENUSE is the greatest side in a triangle!!
And that's where your mistake is!
So you should take the greatest side as C
So in Q3. 7, 24 and 26 are the given numbers
You'll make the smaller two numbers a and b and the greatest number C
Using the Formula you'll solve the left side first which is
{a}^{2}  +  {b}^{2}
Then the right side which is
{c}^{2}

And if both are equal then it is a right triangle otherwise it isn't!
Let
a=7
b=24
c=26


a^2 + b^2
7^2 + 24^2
49 + 576 = 625
GREAT, Now the right side

26^2 = 676
Since they aren't equal it isn't a right angled triangle...

Then let
a=7.5
b=10
c=12.5

7.5^2 + 10^2
= 56.25 + 100
= 156.25



12.5^2 = 156.25

They are EQUAL
Therefore it is a right triangle too

Hopefully I helped
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Johanna wrote the system of equations. 4 x - 3 y = 1, 5 x + 4 y = 9. If the second equation is multiplied by 4, what should the
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Answer:

The first equation must be multiplied by -5 to eliminate x variable by addition

Step-by-step explanation:

4 x - 3 y = 1 (1)

5 x + 4 y = 9 (2)

If the second equation is multiplied by 4

5x+4y=9. ×4

We have,

20x+16y=36 (3)

The first equation should be multiplied by -5 to eliminate x variable by addition

4x-3y=1 × -5

We have

-20x+15y=-5 (4)

Add equation (3) and (4) to eliminate x variable

20x+16y=36

-20x+15y=-5

31y=31

Divide both sides by 31

y=1

Substitute y=1 into equation (1)

4 x - 3 y = 1

4x-3(1)=1

4x-3=1

4x=1+3

4x=4

Divide both sides by 4

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4 0
3 years ago
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SEE ATTACHED
fenix001 [56]
To me personally, the first bit f(g(x)) is easy and the domain is tricky. Let's try explain this.

A function takes an input number and returns an output number depending on the function. Look at f(x) = x+3, if we let the input number be 2 then we say that f(2) = 5. We could do f(π) to give us π+3 or even f(x²) to give us x² +3. The trick is to substitute the input into the function equation.

You have been asked to find f(g(x)). You know f(x) = \frac{1+x}{1-x}. Putting numbers in at this point would be easy (try work out f(2), you'll do it really quick) but you have to put in g(x).

f(g(x)) = \frac{1+g(x)}{1-g(x)}
we also know that g(x) = \frac{x}{1-x} so we can say that
f(g(x)) = \frac{1+ \frac{x}{1-x} }{1- \frac{x}{1-x} } and that is f(g(x)) but the question requires that we simplify it so
\frac{1+ \frac{x}{1-x} }{1- \frac{x}{1-x} }  =  \frac{ \frac{1-x}{1-x} +  \frac{x}{1-x} }{ \frac{1-x}{1-x} - \frac{x}{1-x} } =  \frac{ \frac{1}{1-x} }{ \frac{1-2x}{1-x} } =  \frac{1}{1-2x}

f(g(x)) = \frac{1}{1-2x}

Now for the tricky bit (for me, at least). The domain is the full set of values that you can 'put in to' the function and still get a real value out. So how do we work out what numbers 'break' the function? I like to use the fact that DIVIDING BY ZERO IS IMPOSSIBLE. What value of x can we put into the function to make it so the function is being divided by 0? i.e. 1-2x = 0 solve that and you have a value of x that isn't part of the domain.

This means the domain is all real numbers EXCEPT the solution to that equation. (Because if we put that value into f(g(x)) it's impossible to get a value out.)

[I know this was a lot to read, if you have any questions or don't get anything feel free to message me or leave a comment.]

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Number of spades = 13

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P(spade) = number of spades / total cards in deck

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Step-by-step explanation:

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