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Ainat [17]
3 years ago
6

Can solve this problem please

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
3 0

Answer:

It is C

Step-by-step explanation:

C is true because you can make 12 per hour in the table shown

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Find the length of XY
Pachacha [2.7K]

Since this is a right triangle, you can use the pythagorean theorem, leg^2+leg^2=hypotenuse^2, to find XY. Since we know that 15 and 17 are our legs and that XY is the hypotenuse, we can solve it as such:

15^2+17^2=XY^2\\ 225+289=XY^2\\ 514=XY^2\\ \sqrt{514}=XY

In short, XY is √514, or 22.67 rounded to the hundreths, units long.

7 0
3 years ago
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Solve Systems of Equations Algebracallyy=x+2 y=-3x
Sphinxa [80]

y = x + 2

y = -3x

Do y = -3x in y = x + 2

y = x + 2

-3x = x + 2

-3x - x = 2

-4x = 2

x = -2/4

x = -1/2

Now put x = -1/2 in y = -3x

y = -3x

y = -3.(-1/2)

y = 3/2

5 0
3 years ago
What is the answer? Please answer correctly
Sloan [31]

Answer:

i think its 7t

Step-by-step explanation:

7 0
4 years ago
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Simplify f+g / f-g when f(x)= x-4 / x+9 and g(x)= x-9 / x+4
steposvetlana [31]

f(x)=\dfrac{x-4}{x+9};\ g(x)=\dfrac{x-9}{x+4}\\\\f(x)+g(x)=\dfrac{x-4}{x+9}+\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)+(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2+x^2-9^2}{(x+9)(x+4)}=\dfrac{2x^2-16-81}{(x+9)(x+4)}=\dfrac{2x^2-97}{(x+9)(x+4)}\\\\f(x)-g(x)=\dfrac{x-4}{x+9}-\dfrac{x-9}{x+4}=\dfrac{(x-4)(x+4)-(x-9)(x+9)}{(x+9)(x+4)}\\\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\=\dfrac{x^2-4^2-(x^2-9^2)}{(x+9)(x+4)}=\dfrac{x^2-16-x^2+81}{(x+9)(x+4)}=\dfrac{65}{(x+9)(x+4)}


\dfrac{f+g}{f-g}=(f+g):(f-g)=\dfrac{2x^2-97}{(x+9)(x+4)}:\dfrac{65}{(x+9)(x+4)}\\\\=\dfrac{2x^2-97}{(x+9)(x+4)}\cdot\dfrac{(x+9)(x+4)}{65}\\\\Answer:\ \boxed{\dfrac{f+g}{f-g}=\dfrac{2x^2-97}{65}}

6 0
4 years ago
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A ball is kicked upward with an initial velocity of 32 feet per second. The ball's height, h (in feet), from the ground is model
stellarik [79]

Answer:

1.        t = 0.995 s

2.       h = 15.92  ft

Step-by-step explanation:

First we have to look at the following formula

Vf = Vo + gt

then we work it to clear what we want

Vo + gt = Vf

gt = Vf - Vo

t = (Vf-Vo)/g

Now we have to complete the formula with the real data

Vo = 32 ft/s      as the statement says

Vf = 0     because when it reaches its maximum point it will stop before starting to lower

g = -32,16 ft/s²        it is a known constant, that we use it with the negative sign because it is in the opposite direction to ours

t = (0 ft/s - 32 ft/s) / -32,16 ft/s²

we solve and ...

t = 0.995 s

Now we will implement this result in the following formula to get the height at that time

h = (Vo - Vf) *t /2

h = (32 ft/s - 0 ft/s) * 0.995 s / 2

h = 32 ft/s * 0.995 s/2

h = 31.84 ft / 2

h = 15.92  ft

4 0
3 years ago
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