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Ira Lisetskai [31]
3 years ago
13

the length of a rectangle is 7 inches more than its width.if the perimeter of the rectangle is 66 inches,find its dimensions.

Mathematics
2 answers:
11111nata11111 [884]3 years ago
5 0
P=66
L=7+w
2L+2W=p
2(7+w)+2w=66
14+2w+2w=66
4w=52
w=13
L=7+w
L=7+(13)
L=20 & w=13
Ronch [10]3 years ago
4 0
Perimeter = 2 times the width + 2 times the length
                = 2(w + 7) + 2w     where w = width

2(w + 7) + 2w = 66

4w + 14 = 66
4w = 66-14 = 52
w = 52/4 = 13

length = 13 + 7 = 20 

width = 13 and length = 20

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Evaluate the expression below using the properties of operations. Show your steps.−36 ÷ 14 ⋅ (−18) ⋅ (−3) ÷ 6
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Answer:

(-162)/7 or -23 1/7 as mixed fraction

Step-by-step explanation:

Simplify the following:

(-36)/14 (-18) (-3)/6

Hint: | Express (-36)/14 (-18) (-3)/6 as a single fraction.

(-36)/14 (-18) (-3)/6 = (-36 (-18) (-3))/(14×6):

(-36 (-18) (-3))/(14×6)

Hint: | In (-36 (-18) (-3))/(14×6), divide -18 in the numerator by 6 in the denominator.

(-18)/6 = (6 (-3))/6 = -3:

(-36-3 (-3))/14

Hint: | In (-36 (-3) (-3))/14, the numbers -36 in the numerator and 14 in the denominator have gcd greater than one.

The gcd of -36 and 14 is 2, so (-36 (-3) (-3))/14 = ((2 (-18)) (-3) (-3))/(2×7) = 2/2×(-18 (-3) (-3))/7 = (-18 (-3) (-3))/7:

(-18 (-3) (-3))/7

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(54 (-3))/7

Hint: | Multiply 54 and -3 together.

54 (-3) = -162:

Answer:  (-162)/7

3 0
4 years ago
P³ = 1/8 please help me if anybody can .
ivanzaharov [21]

Answer:

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Step-by-step explanation:

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\textsf{Apply exponent rule} \quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}:

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\textsf{Apply exponent rule} \quad 1^a=1:

\implies p= \dfrac{1}{8^{\frac{1}{3}}}

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\implies p= \dfrac{1}{(2^3)^{\frac{1}{3}}}

\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:

\implies p= \dfrac{1}{2^{(3 \cdot \frac{1}{3})}}

Simplify:

\implies p= \dfrac{1}{2^{\frac{3}{3}}}

\implies p= \dfrac{1}{2^{1}}

\implies p= \dfrac{1}{2}

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1 year ago
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Liono4ka [1.6K]

Answer:

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Step-by-step explanation:

We proceed to derive each expression by rule of chain. Let be w = \ln (x+2\cdot y + 3\cdot z), x = r^{2}+t^{2}, y = s^{2}-t^{2} and z = r^{2}+s^{2}:

\frac{dw}{dr} = \frac{\frac{dx}{dr}+2\cdot \frac{dy}{dr} +3\cdot \frac{dz}{dr}  }{x+2\cdot y + 3\cdot z}

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\frac{dy}{dr} = 0

\frac{dz}{dr} = 2\cdot r

\frac{dw}{dr} = \frac{8\cdot r}{(r^{2}+t^{2})+2\cdot (s^{2}-t^{2})+3\cdot (r^{2}+s^{2})}

\frac{dw}{dr} = \frac{8\cdot r}{4\cdot r^{2}-t^{2}+5\cdot s^{2}} (1)

\frac{dw}{ds} = \frac{\frac{dx}{ds}+2\cdot \frac{dy}{ds} +3\cdot \frac{dz}{ds}  }{x+2\cdot y + 3\cdot z}

\frac{dx}{ds} = 0

\frac{dy}{ds} = 2\cdot s

\frac{dz}{ds} = 2\cdot s

\frac{dw}{ds} = \frac{10\cdot s}{(r^{2}+t^{2})+2\cdot (s^{2}-t^{2})+3\cdot (r^{2}+s^{2})}

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\frac{dw}{dt} = \frac{\frac{dx}{dt}+2\cdot \frac{dy}{dt} +3\cdot \frac{dz}{dt}  }{x+2\cdot y + 3\cdot z}

\frac{dx}{dt} = 2\cdot t

\frac{dy}{dt} = -2\cdot t

\frac{dz}{dt} = 0

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7 0
3 years ago
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