What will the remainder over the divisor?

Nat2105 [25]

ABC and HJK are congruent and the corresponding sides and angles are equal. Side HJ is congruent with side AB so HJ is also 3.5cm in length.

8 0

3 years ago

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Hilda was simplifying some numerical expressions and made each of the following sequences of calculations. Name the mathematical

Galina-37 [17]

Answer:

Problem 1:

5 · (−4/3) · (2/5)

= (−4/3) · 5 · (2/5) Conmutative Property of Multiplication

= (−4/3) · (5 · (2/5)) Associative Property of Multiplication

= (−4/3) · (2/1) = −8/3 = −2*2/3 Multiplied fractions and extracted common factor

Problem 2:

17 + 29 + 3+ 1

= 17 + 3 + 29 + 1 Conmutative Property of Addition

= (17 + 3) + (29 + 1) Associative Property of Addition

= 20 + 30 Added groups

= 50 Added terms

Step-by-step explanation:

For the Problem 1:

In the first step, Hilda applied the Conmutative Property of Multiplication, because she changed the order of the numbers in the product

In the second step, she applied the Associative Property of Multiplication, because she agrouped the product of 5 and 2/5 to perform it sepparately

In the third step, she calculated the product of the fractions -4/3 and 2/1, then she extracted 2 as a common factor to express the fraction as -2*2/3

For the Problem 2:

In the first step, Hilda applied the Conmutative Property of Addition, because she changed the order of the numbers in the sum

In the second step, she applied the Associative Property of Addition, because she associated the addition of 17 and 3 and the addition of 29 and 1, to calculate them in groups.

4 0

3 years ago

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PLEASE HELP I WILL GIVE BRAINLEST

nadya68 [22]

OK. All the work is in the attached drawing.

Please spend enough time with it to understand what was done and how, so that you can do the next problem like this one completely on your own.

Good luck !

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Here's a picky pickypoint to think about concerning the "cost". I'm not quite sure what to do about this:

The initial, one-time $55 is a "deposit". If she doesn't smash the bike or steal it, she gets the $55 back when she returns the bike.

So if she eventually gets the $55 back, is it a cost ? ? I don't know how to think about it.

She does need to give them the $55 at the beginning, just to take the bike out. So she has to have it with her and give it to them temporarily, even though she'll get it back, and she'll still have it when she returns home.

So here's the story:

-- While she HAS the bike, the graph correctly shows all the money she needs, in order to get the bike, and keep it for however long she keeps it.

-- Finally, at the end of the week, after she returns the bike and gets her $55 back, the line and everything on the graph will shift down by $55. The line will start at zero, and the little red ordered pairs will also shift straight down and still be on the line.

To put it one more way:

-- While Jen has the bike, the y-intercept of the graph is $55.

-- After she returns the bike in good condition, the y-intercept is zero.