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lisabon 2012 [21]
3 years ago
14

watermelons cost $.39 per pound at a local market. Kent’s watermelon cost between $4.00 and $5.00. What are the possible weights

of his watermelon?
Mathematics
1 answer:
faust18 [17]3 years ago
5 0

4/.39=10.26 lbs.


5/.39=12.83 lbs.

The melons are between <u>10.26 & 12.83 lbs</u>.

:)

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The base of an isosceles triangle is 5 and it's perimeter is 11. The base of a similar isosceles triangle is 10.What is the peri
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Answer: the perimeter of the larger triangle is 22 units

Step-by-step explanation:

The diagram of both triangles is shown in the attached photo. The isosceles triangles are triangle ABC and triangle DEF

Perimeter of triangle ABC = 11

For an isosceles triangle, the other two sides are equal. If one equal side is x, then the sum of the two equal sides is x+x = 2x

Perimeter of triangle = sum of the lengths of the 3 sides. Therefore,

11 = 5 + 2x

2x = 11-5 = 6

x = 6/2 = 3

The base of the similar triangle is 10. It means that it is 2 times the base of the smaller triangle. That means the scale factor is 2. Therefore,

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3×2 + 3×2 + 5×2 = 6+6+10 = 22 units

5 0
3 years ago
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What is sin45°*sin30°*cos60°
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\frac{\sqrt{2} }{8}

Step-by-step explanation:

Using the exact values of the given trig functions, then

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It took Joe 3 hours swimming at a speed of 12 miles per hour to swim along the Charles River. If Jennifer is swimming the same p
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Answer:

4 hours

Step-by-step explanation:

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7 0
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Answer:

Therefore, the inverse of given matrix is

=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}

Step-by-step explanation:

The inverse of a square matrix A is A^{-1} such that

A A^{-1}=I where I is the identity matrix.

Consider, A = \left[\begin{array}{ccc}4&3\\3&6\end{array}\right]

\mathrm{Matrix\:can\:only\:be\:inverted\:if\:it\:is\:non-singular,\:that\:is:}

\det \begin{pmatrix}4&3 \\3&6\end{pmatrix}\ne 0

\mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}:\quad \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\frac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}

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\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}\:=\:ad-bc

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Therefore, the inverse of given matrix is

=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}

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Answer:

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hope it helps.

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