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xeze [42]
2 years ago
5

Verify X + y equal to y + X when:x=-8/9and y=-3/5​

Mathematics
1 answer:
atroni [7]2 years ago
4 0

Given:

x=-\dfrac{8}{9},\ y=-\dfrac{3}{5}

To verify:

x+y=y+x

Solution:

We need to verify,

x+y=y+x

Taking left hand side, we get

LHS=x+y

After substituting the values, we get

LHS=-\dfrac{8}{9}+\left(-\dfrac{3}{5}\right)

LHS=-\dfrac{8}{9}-\dfrac{3}{5}

LHS=\dfrac{-40-27}{45}

LHS=\dfrac{-67}{45}

Now,

Taking right hand side, we get

RHS=y+x

After substituting the values, we get

RHS=-\dfrac{3}{5}+\left(-\dfrac{8}{9}\right)

RHS=-\dfrac{3}{5}-\dfrac{8}{9}

RHS=\dfrac{-27-40}{45}

RHS=\dfrac{-67}{45}

Clearly,

LHS=RHS

Hence verified.

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Hey can u answer my math question look atbthe picture to see<br> the pic.​
Anna35 [415]

Answer:

4 3/4 yards

Step-by-step explanation:

5 1/4 + 2 1/2 = 7 3/4

(1/2 is equal to 2/4)

12 1/2 - 7 3/4 = 4 3/4

Hope this helps, mark me brainliest if it does! :)

4 0
2 years ago
Read 2 more answers
HELP<br><br>jusko ang hirap<br><br><br><br>❗NONSENSE = REPORTED❗​
irga5000 [103]

Answer:

1.A

2.B

3.A

4. D

5.C

Step-by-step explanation:

make me brainlest :>>

4 0
2 years ago
Please help immediately
Mila [183]
Add all of the scores together which gives you

550+-1500+-2200+400+-600+-300

= -3650

Hope this helps! (:
5 0
3 years ago
Read 2 more answers
What are the restrictions for a? 2a^2+a-15/5a^2+16a+3
koban [17]

ANSWER


The restrictions are

a\ne -3,a\ne -\frac{1}{5}


EXPLANATION


We were given the rational function,


\frac{2a^2+a-15}{5a^2+16a+3}


The function is defined for all values of a, except



5a^2+16a+3=0


This has become a quadratic trinomial, so we need to split the middle term.


We do that by multiplying the coefficient of x^2 which is 5 by the constant term which is 3. This gives us 15.


The factors of 15 that adds up to 16 are 1 and 15.


We use these factors to split the middle term.




5a^2+15a+a+3=0


We now factor to get,


5a(a+3)+1(a+3)=0


We factor further to get,


(a+3)(5a+1)=0



This implies that,


(a+3)=0,(5a+1)=0


This gives


a=-3,a=-\frac{1}{5}


These are the restrictions.





8 0
3 years ago
A rectangle has length 4 inches and width 2 inches. If the length and width of the rectangle are each reduced by 50 percent, by
exis [7]

Answer: 25%

Step-by-step explanation:

Given: A rectangle has length 4 inches and width 2 inches.

Area = Length x width

= 4 inches x  2 inches

= 8 square inches

If length is reduced by 50% , then length =0.5\times4=2\text{ inches}

If width is reduced by 50% , then length =0.5\times2=1\text{ inch}

Reduced area = (reduced length) x (reduced width)

= 2 inches x 1 inch

= 2 square inches

The percentage of the area of the rectangle be reduced :-

\dfrac{\text{Reduced area}}{\text{Original area}}\times100 \\\\=\dfrac{2}{8}\times100\% \\\\=25\%

Hence, the percent of the area of the rectangle be reduced = 25%

6 0
3 years ago
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