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irina [24]
3 years ago
12

How do I find the value of e any help please​

Mathematics
1 answer:
anygoal [31]3 years ago
4 0

Answer:

97

Step-by-step explanation:

360-117-146 is 97

Gimmi brainest answer :)

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Let f(x) = x2 − 16. Find f−1(x)
Yanka [14]

Answer:

f^{-1}(x)=\pm\sqrt{x+16}

Step-by-step explanation:

f(x)=x^2-16

To find the inverse of this function, temporarily the f(x) with a y (or whatever else you want) and then solve for the x.

y=x^2-16\\y+16=x^2\\x=\pm\sqrt{y+16}

Now, you can swap the variables again and write the function.

f^{-1}(x)=\pm\sqrt{x+16}

3 0
3 years ago
Evaluate x(y+3)/(3+y)z for x=12, y=1, and z=6.<br><br> A) 2<br><br> B) 5<br><br> C) 8<br><br> D) 6
fgiga [73]

Answer:

2

Step-by-step explanation:

We take the equation

x(y+3)/(3+y)z

and substitute the values for each individual variable in the problem. It looks like this:

12(1+3)/(3+1)6

Now we can solve the equation.

When solved, it equals 2.

5 0
3 years ago
Read 2 more answers
Please help me with this
Alisiya [41]
Woahhhh lol one sec
8 0
3 years ago
Ramona wants to buy cheese pizza for her family and friends gathering. She stop by nine wonder pizza and figures out that they h
Kryger [21]

Answer:

Large cheese

Step-by-step explanation:

The large cheese because it's two more dollars than the medium and she has a lot of people gathering

6 0
3 years ago
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The center of a hyperbola is located at the origin. One focus is located at (−50, 0) and its associated directrix is represented
leva [86]

The equation of the hyperbola is : \frac{x^{2}}{48^2}  - \frac{y^{2}}{14^2}  = 1

The center of a hyperbola is located at the origin that means at (0, 0) and one of the focus is at (-50, 0)

As both center and the focus are lying on the x-axis, so the hyperbola is a horizontal hyperbola and the standard equation of horizontal hyperbola when center is at origin: \frac{x^{2}}{a^{2}}  - \frac{y^{2}}{b^{2}}    = 1

The distance from center to focus is 'c' and here focus is at (-50,0)

So, c= 50

Now if the distance from center to the directrix line is 'd', then

d= \frac{a^{2}}{c}

Here the directrix line is given as : x= 2304/50

Thus, \frac{a^{2}}{c}  = \frac{2304}{50}

⇒ \frac{a^{2}}{50}  = \frac{2304}{50}

⇒ a² = 2304

⇒ a = √2304 = 48

For hyperbola, b² = c² - a²

⇒ b² = 50² - 48² (By plugging c=50 and a = 48)

⇒ b² = 2500 - 2304

⇒ b² = 196

⇒ b = √196 = 14

So, the equation of the hyperbola is : \frac{x^{2}}{48^2}  - \frac{y^{2}}{14^2}  = 1

5 0
3 years ago
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