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Jlenok [28]
3 years ago
7

A fish tank contains as 1/2 many goldfish as mollies, and there are 6 more guppies than goldfish. If the total combined number o

f fish in the tank is 22, how many mollies are in the tank?
Mathematics
2 answers:
Bezzdna [24]3 years ago
8 0

Answer:17

Step-by-step explanation:

defon3 years ago
4 0
17, because you divide 22 in half and add 6 because there are half as many mollies and there are 6 more guppies then goldfish.
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I have to find the area
Alina [70]
The formula for finding the area of a triangle is: base*height*1/2
So, if we use this information....
4.2*5.5*1/2=11.55
Area=11.55
4 0
3 years ago
Read 2 more answers
When the two equations are graphed on a coordinate plane, they intersect at two points.
Law Incorporation [45]

Answer:

(0,4) and (-2,0)

Step-by-step explanation:

y=3x²+4x+4

y = -2x +4

3x²+4x+4 = - 2x +4

3x²+6x=0

3x(x+2)=0

x=0, x= -2

x=0, y=-2x+4= -2*0 + 4 =4,   (0,4)

x= - 2, y = 2x + 4= 2*(-2) +4=0,    (-2,0)

3 0
3 years ago
Jeremy is going on rides at a carnival he starts the rides a 12:30 by 12:45 he had gone on 1/5 of the carnival rides at the rate
MArishka [77]

Answer:

1.25 hours or 1 hour and 15 minutes

Step-by-step explanation:

12:45 - 12:30 = :15

15 minutes equates to 1/5 of the rides

So, we multiply 15 by 5 to find the total time taken to ride all the rides.

15 x 5 = 75

75/60 = 1.25

It would take him 1.25 hours to ride all the rides at the carnival or 1 hour and 15 minutes.

8 0
3 years ago
Find the nth taylor polynomial for the function, centered at c. f(x) = ln(x), n = 4, c = 5
masha68 [24]

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Given:

f(x) = ln(x)

n = 4

c = 3

nth Taylor polynomial for the function, centered at c

The Taylor series for f(x) = ln x centered at 5 is:

P_{n}(x)=f(c)+\frac{f^{'} (c)}{1!}(x-c)+  \frac{f^{''} (c)}{2!}(x-c)^{2} +\frac{f^{'''} (c)}{3!}(x-c)^{3}+.....+\frac{f^{n} (c)}{n!}(x-c)^{n}

Since, c = 5 so,

P_{4}(x)=f(5)+\frac{f^{'} (5)}{1!}(x-5)+  \frac{f^{''} (5)}{2!}(x-5)^{2} +\frac{f^{'''} (5)}{3!}(x-5)^{3}+.....+\frac{f^{n} (5)}{n!}(x-5)^{n}

Now

f(5) = ln 5

f'(x) = 1/x ⇒ f'(5) = 1/5

f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25

f'''(x) = 2/x³  ⇒ f'''(5) = 2/5³ = 2/125

f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625

So Taylor polynomial for n = 4 is:

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Hence,

The nth taylor polynomial for the given function is

P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴

Find out more information about nth taylor polynomial here

brainly.com/question/28196765

#SPJ4

3 0
2 years ago
What is the slope of the line that passes through the points (5,3) and (-4,1)​
wariber [46]

Answer:

\frac{2}{9}

Step-by-step explanation:

Formula to find the slope:\frac{y_{2}-y_1}{x_2-x_1}

Notice it doesn't matter for which point being y2/x2 or y1/x1.

Just to make our life easier, I choose (5,3) as our second point and (-4,1) as our first point.

\frac{3-1}{5-(-4)}

=\frac{2}{9}

5 0
3 years ago
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