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Luba_88 [7]
3 years ago
6

if doris paid $5.28 for 4.4 pounds of swiss cheese what was the the price of 1 pound of swiss cheese?

Mathematics
2 answers:
victus00 [196]3 years ago
8 0

5.28/ 4.4 = 1.2 per pound

Nutka1998 [239]3 years ago
6 0
I think you multiply
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The figure is transformed as shown in the diagram. Describe the transformation. A) dilation,then reflection B) reflection, then
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A! The polygon was dilated, then reflection vertically!

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The distance by road from town A to town B is 257 km. What is 50% of that distance?
Marina86 [1]

Answer: 128.5km

Step-by-step explanation:

Since we are given the information that the distance by road from town A to town B is 257 km. To get 50% of the distance, we simply have to multiply the distance given by 50%. This will be:

= 50% × 257km

= 50/100 × 257km

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= 128.5km

Therefore, 50% of the distance is 128.5km.

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To the nearest tenth, what is the length of the hypotenuse of an isosceles right triangle with a leg of 7 3 inches?
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3 years ago
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1/6(a+12)= -4 please help because I don’t understand this
iogann1982 [59]

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a=-36

Step-by-step explanation:

1/6(a+12)=-4

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1/6a+2=-4

2) Subtract 2 from both sides:

1/6a=-6

3) Multiply both sides by the reciprocal of 1/6: (6/1)

a=-36

Let me know if you have any confusion on how I reached the solution :)

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2 years ago
If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?
Alisiya [41]
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:

\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0

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
S^*=2(k-2)

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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