Joanna is carrying her books to school in her backpack. Her library book weighs \dfrac{3}{8} 8 3 ​ start fraction, 3, divided by, 8, end fraction of a kilogram, her science book weighs \dfrac{11}{8} 8 11 ​ start fraction, 11, divided by, 8, end fraction kilograms, and her math book weighs \dfrac{5}{8} 8 5 ​ start fraction, 5, divided by, 8, end fraction kilograms.

no, the equation would be invalid since they are both not equal.

step-by-step explanation:

here you got

4y-6+4y 2-9

first, you would simplify the equation on the form.

4y-6+4y 2-9

then, move all the terms that contains a y to the left and all the other terms to the right.

+4y-4y+-9-6

finally, you will simplify the left and the right side of the equation.

y=-3

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.