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Mazyrski [523]
3 years ago
13

A rectangle has an area of 56 square centimeters it's length is 8 centimeters what's the width

Mathematics
2 answers:
mario62 [17]3 years ago
7 0
You take 56 and divide by 8 (area is the length times width) and get 7
Kamila [148]3 years ago
6 0
A= L × W
So W = A ÷ L
So W = 56 ÷ 8
So W =  7 cm
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A picture measuring 2.5" high by 6" wide is to be enlarged so that the width is now 18 inches. How tall will the picture be?
Usimov [2.4K]
7.5" if the width is now 18 that means its value increased by 3 so multiply 2.5 by 3 and its = to 7.5
6 0
3 years ago
Read 2 more answers
Samantha bought 26 total fish. Each goldfish cost $3 and each guppy costs $4. She spent $86 total. How many guppies did she buy?
Alex787 [66]

Answer:

goldfish is 15 and guppies is 11

Step-by-step explanation:

Given data

let the number of goldfish be x

and the number of guppies be y

so

x+y= 25----------1

and

3x+4y= 86-------2

solve 1 and 2

x+y= 25

3x+4y= 86

from 1

x= 25-y

put this in 2

3(25-y)+4y= 86

75-3y+4y= 86

75+y= 86

y= 86-75

y=11

put y= 11 in 1

x+11= 26

x= 26-11

x= 15

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5 0
3 years ago
Use the chain rule to differentiate the following function.<br><br>f(x)=(5x+sin^3(x)+sinx^3)^3
Bumek [7]

f(x)=(5x+\sin^3(x)+\sin x^3)^3\\\frac{df(x)}{dx}=3(5x+\sin^3(x)+\sin x^3)^2\cdot (5+3\sin^2x\cos x+3x^2\cos x^3)

4 0
3 years ago
Which answer describes the simpler problems that could be used to solve this story problem? Jackson read some books over the sum
Fittoniya [83]
The answer should be c
5 0
3 years ago
A triangle has two constant sides of length 5 feet and 7 feet. The angle between these two sides is increasing at a rate of 0.9
o-na [289]

Answer:12.74 ft^2/s

Step-by-step explanation:

Given

Two sides of triangle of sides 5 ft and 7 ft

and angle between them is increasing at a rate of 0.9 radians per second

let \thetais the angle between them thus

Area of triangle when two sides and angle between them is given

A=\frac{ab\sin C}{2}

A=\frac{5\times 7\times \sin \theta }{2}

Differentiate w.r.t time

\frac{\mathrm{d} A}{\mathrm{d} t}=\frac{35\cos theta }{2}\times \frac{\mathrm{d} \theta }{\mathrm{d} t}

at \theta =\frac{\pi }{5}

\frac{\mathrm{d} A}{\mathrm{d} t}=\frac{35\times cos(\frac{\pi }{5})}{2}\times 0.9

\frac{\mathrm{d} A}{\mathrm{d} t}=12.74 ft^2/s

4 0
3 years ago
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