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2 years ago

Gary divides 48.026 by 6.49 using the work shown. Is Gary correct or incorrect? Explain your answer in full sentences.

Mathematics

1 answer:

Likurg_2 [28]2 years ago

3 0

Answer:

Here, we are required to judge judge Gary's calculation as portrayed in the question above.

Gary's calculation is correct according to the calculation.

Step-by-step explanation:

The question asks that 48.026 is divided by 6.49.

By first considering the whole numbers on the left hand side of the decimal point in each number,

We can conclude that 48 divided by 6 equals 8, i.e 48/6=8.

However, the calculation becomes much easier by shifting the decimal points in both numbers two spaces to the right as indicated in the diagram (i.e multiply both numbers by 100).

This leaves us with 4802.6 to be divided by 649.

Dividing 4802.6 by 649 yields 7 with a remainder of 259.6 (this is so because 7 × 649 = 4543).

259.6 can have it's decimal point shifted one space right to yield 2596.

And, 2596/649 = 4.

The condition above is hypothetical, and there's a need for the decimal point to be shifted one space back to the left, which leaves us with 0.4.

Ultimately, 48.026/6.49 = 7.4.

Therefore, Gary's calculation is correct using the work shown in the question above.

brainly.com/question/20915806&ved=2ahUKEwiRtIygpIbzAhUN8xQKHd9LACkQjjh6BAgnEAE&usg=AOvVaw2ymwQ-_WS0EVfUGoAy5WB_

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3 years ago

If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?

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If $k$ is odd, then

$\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor$

while if $k$ is even, then the sum would be

$\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4$

The latter case is easier to solve:

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which means $k=6$ .

In the odd case, instead of considering the above equation we can consider the partial sums. If $k$ is odd, then the sum of the even integers between 1 and $k$ would be

$S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)$

Now consider the partial sum up to the second-to-last term,

$S^*=2+4+6+\cdots+(k-5)+(k-3)$

Subtracting this from the previous partial sum, we have

$S-S^*=k-1$

We're given that the sums must add to $2k$ , which means

$S=2k$
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But taking the differences now yields

$S-S^*=2k-2(k-2)=4$

and there is only one $k$ for which $k-1=4$ ; namely, $k=5$ . However, the sum of the even integers between 1 and 5 is $2+4=6$ , whereas $2k=10\neq6$ . So there are no solutions to this over the odd integers.

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3 years ago

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weeeeeb [17]

Answer:

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Step-by-step explanation:

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subtract x from both sides

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3 years ago