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patriot [66]
3 years ago
5

A rectangular field is 125 meters long and 75 meters wide. Give the length and width of another rectangular field that has the s

ame perimeter but a smaller area.
Mathematics
1 answer:
mario62 [17]3 years ago
5 0

Answer:

150 m and 50 m

Step-by-step explanation:

l= 125 m

w= 75 m

P= 2(125+75)= 400 m

A= 125*75= 9375 m²

l1= 125+25= 150 m

w1= 75-25= 50 m

A1= 150*50= 7500 m² smaller area with same perimeter

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salantis [7]
That would be DF is congruent to DE     ( first choice)

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4 years ago
Jason’s mom gave him 1,000 for his birthday. He put the money in the bank and decides to withdraw 25 per week.
Goshia [24]

Well 1000 for his birthday then subtract 25 each week, w, should be 1000-25w if i am not mistaken.

6 0
4 years ago
Jessie designed a sculpture that is shaped like a circle. The circumference of the sculpture is 10π meters. Which measurement is
BlackZzzverrR [31]

Answer:

25πm²

Step-by-step explanation:

Circumference of the circular sculpture = πd

d is the diameter

Given

Circumference = 10π metres

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Hence the measurement closest to the area of the sculpture in square meters is 25πm²

6 0
3 years ago
Sin α = 21/29, α lies in quadrant II, and cos β = 15/17, β lies in quadrant I Find sin (α - β).
Sever21 [200]
\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta

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Since \alpha lies in quadrant II, we have \cos\alpha, so

\cos\alpha=-\sqrt{\dfrac{400}{841}}=-\dfrac{20}{29}

\cos\beta=\dfrac{15}{17}\implies\sin^2\beta=1-\cos^2\beta=\dfrac{64}{289}

\beta lies in quadrant I, so \sin\beta>0 and

\sin\beta=\sqrt{\dfrac{64}{289}}=\dfrac8{17}

So

\sin(\alpha-\beta)=\dfrac{21}{29}\dfrac{15}{17}-\left(-\dfrac{20}{29}\right)\dfrac8{17}=\dfrac{475}{493}
8 0
3 years ago
99 POINTS!!!! PLEASE HELP!! WILL MARK BRAINLIST!!
RUDIKE [14]

The answer is reduction. This can be found by the size and number differences in the two quadrilaterals.

8 0
4 years ago
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