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Aleks [24]
3 years ago
11

Rey feeds his dog 2/5 of a can of dog food in the morning and 4/5 of a can in the evening. How many cans of dog food will Rey ne

ed in order to feed his dog for five days?
Mathematics
1 answer:
QveST [7]3 years ago
6 0

Answer:

6 cans total if im correct

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F(x)=2x^2+3 and g(x)=x^2-7 find (f+g)(x)
defon

Answer:

3x^2 -4

Step-by-step explanation:

F(x)=2x^2+3 and g(x)=x^2-7

(f+g)(x) =

We add the two functions together

(f+g)(x) =2x^2+3+x^2-7

I like to line them up vertically

(f+g)(x) =

2x^2 +3

+x^2  -7

------------------

3x^2 -4

3 0
3 years ago
Simplified form of this expression 5z-8-4z-3z+6
kvasek [131]

Answer:

-2z-2

Step-by-step explanation:

5z-8-4z-3z+6

5z-4z-3z-8+6

-2z-2

7 0
3 years ago
Read 2 more answers
The amount $180.00 is what percent greater than $135.00?
svp [43]
It would be: 180/135 * 100 = 36/27 * 100
= 4/3 * 100
= 400/3 = 133.33

It means, 180 is 133.33% of 135. So, the difference would be: 133.33 - 100 = 33.33%

In short, Your Final Answer would be 33.33%

Hope this helps!
5 0
3 years ago
Read 2 more answers
Solve the following equation.<br> x^2 + 6x − 40 = 0
jonny [76]

Answer:

x = 4 and -10

Step-by-step explanation:

Given the equation

x²+6x-40= 0

The common factors to use for the factorisation are +10 and -4

x²+(10x-4x)-40=0

(x²+10x)-(4x-40) = 0

x(x+10)-4(x+10) = 0

(x-4)(x+10) = 0

x-4 =0 and x+10 = 0

x = 4 and x = -10

8 0
3 years ago
the Hog's head pub decide to make mixture thst is 50% butterbeer. how much of an 80% butterbeer solution should mixed with 20 ga
evablogger [386]
There are 20 gallons of 20% butterbeer solution.
Assume there are x gallons of 80% solution.

So total amount of the solution will be (x+20) gallons
The final solution is 50%.

So, we can set up the equation as:

0.2(20)+0.8(x)=0.5(x+20) \\ \\ 4+0.8x=0.5x+10 \\ \\ 0.8x-0.5x=10-4 \\ \\ 0.3x=6 x=20.

Thus, there will be 20 gallons of 80% butterbeer solution that is needed to be mixed with 20 gallons of 20% butterbeer solution. 

4 0
3 years ago
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