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erastovalidia [21]
2 years ago
6

Can anyone help! Struggling on this question!!

Mathematics
2 answers:
prohojiy [21]2 years ago
4 0

Answer:

\frac{1}{10}

Step-by-step explanation:

note that in terms of the given fraction

1 = \frac{10}{10} , then

\frac{9}{10} + \frac{1}{10} = \frac{10}{10} = 1

shusha [124]2 years ago
4 0

Answer:

\frac{9}{10}  + x = 1 \\ x = 1 -  \frac{9}{10}  \\ x =  \frac{(10 - 9)}{10}  \\  \boxed{x =  \frac{1}{10} } \: is \: the \: required \: number

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  • See below

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Select correct answer
Sergio [31]

Answer:

The values of p in the equation are 0 and 6

Step-by-step explanation:

First, you have to make the denominators the same. to do that, first factor 2p^2-7p-4 = \left(2p+1\right)\left(p-4\right)2p

2

−7p−4=(2p+1)(p−4)

So then the equation looks like:

\frac{p}{2p+1}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{5}{p-4}

2p+1

p

−

(2p+1)(p−4)

2p

2

+5

=−

p−4

5

To make the denominators equal, multiply 2p+1 with p-4 and p-4 with 2p+1:

\frac{p^2-4p}{(2p+1)(p-4)}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{10p+5}{(p-4)(2p+1)}

(2p+1)(p−4)

p

2

−4p

−

(2p+1)(p−4)

2p

2

+5

=−

(p−4)(2p+1)

10p+5

Since, this has an equal sign we 'get rid of' or 'forget' the denominator and only solve the numerator.

(p^2-4p)-(2p^2+5)=-(10p+5)(p

2

−4p)−(2p

2

+5)=−(10p+5)

Now, solve like a normal equation. Solve (p^2-4p)-(2p^2+5)(p

2

−4p)−(2p

2

+5) first:

(p^2-4p)-(2p^2+5)=-p^2-4p-5(p

2

−4p)−(2p

2

+5)=−p

2

−4p−5

-p^2-4p-5=-10p+5−p

2

−4p−5=−10p+5

Combine like terms:

-p^2-4p+0=-10p−p

2

−4p+0=−10p

-p^2+6p=0−p

2

+6p=0

Factor:

p=0, p=6p

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