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alukav5142 [94]
2 years ago
13

Can anyone help me with this question please I will mark as a brainliest.

Mathematics
1 answer:
Scorpion4ik [409]2 years ago
8 0

Answer:

I think it is A.

Step-by-step explanation:

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Please help quickly! will give brainliest! no explanation is required. just a correct answer :)
myrzilka [38]

Answer:

A

Step-by-step explanation:

8 0
3 years ago
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Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar.
guajiro [1.7K]

Answer:

5 - x

Step-by-step explanation:

Given:

f(x)=25-x^2

g(x)=x+5

\begin{aligned}\left(\dfrac{f}{g}\right)(x) & = \dfrac{f(x)}{g(x)}\\\\ & = \dfrac{25-x^2}{x+5}\\\\& = \dfrac{(5-x)(5+x)}{(x+5)}\\\\& = \dfrac{(5-x)(x+5)}{(x+5)}\\\\& = 5-x\end{aligned}

8 0
2 years ago
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Prove that the gregorian calendar repeats itself every 400 years.
Setler79 [48]

The calendar obviously has an integral number of years and months in 400 years. If it has an integral number of weeks, then it will repeat itself after that time. The rules of the calendar eliminate a leap year in 3 out of the four century years, so there are 97 leap years in 400 years. The number of excess days of the week in 400 years can be found by ...

(303·365) mod 7 + (97·366) mod 7 = (2·1 + 6·2) mod 7 = 14 mod 7 = 0

Thus, there are also an integral number of weeks in 400 years.

The first day of the week is the same at the start of every 400-year interval, so the calendar repeats every 400 years.

3 0
3 years ago
a circle has a diameter of 10cm what is the best proximation of this area use 3.14 to approximate for pi
hammer [34]
Hello!

The answer would be 75.8^{cm2}

Hope this helped!
4 0
3 years ago
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Find an equation in standard form for the hyperbola with vertices at (0, ±3) and foci at (0, ±7)
Nitella [24]

The equation of a hyperbola is:

(x – h)^2 / a^2 - (y – k)^2 / b^2 = 1

 

So what we have to do is to look for the values of the variables:

<span>For the given hyperbola : center (h, k) = (0, 0)
a = 3(distance from center to vertices)
a^2 = 9</span>

<span>
c = 7 (distance from center to vertices; given from the foci)
c^2 = 49</span>

 

<span>By the hypotenuse formula:
c^2 = a^2 + b^2
b^2 = c^2 - a^2 </span>

<span>b^2 =  49 – 9</span>

<span>b^2  = 40

</span>

Therefore the equation of the hyperbola is:

<span>(x^2 / 9) – (y^2 / 40) = 1</span>

5 0
3 years ago
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