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GREYUIT [131]
3 years ago
5

The perimeter of a rectangle is 128 cm, and the ratio of its sides is 5: 3. Find the area of ​​a rectangle

Mathematics
2 answers:
blondinia [14]3 years ago
7 0

Answer:

Area of the rectangle =15 cm^2

Step-by-step explanation:

Perimeter of the rectangle= 128 cm

Ratio of length to width is:

\frac{Length}{Width}= \frac{5}{3} \\\\Length=5 cm\\\\Width=3 cm

Area of the rectangle= Length*Width

=5cm*3cm\\\\=15cm^2

Area of the rectangle =15 cm^2

Akimi4 [234]3 years ago
4 0

Answer:

960 cm^2

Step-by-step explanation:

length: width

5  :3

Multiply by x

5x  ; 3x

The perimeter of a rectangle is found by

P =2(l+w)

128 = 2 (5x+3x)

Combine like terms

128 = 2(8x)

128 = 16x

Divide each side by 16

128/16 = 16x/16

8 =x

length : width

5*8:  3*8

40 : 24

We need to find the area

A = l*w

  = 40*24

   =960

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3 years ago
3x^2+48=0, solve the equation
MAXImum [283]

Answer:

16 + x2 = 0     (16)

I THINK

Step-by-step explanation: Simplifying

3x2 + 48 = 0

Reorder the terms:

48 + 3x2 = 0

Solving

48 + 3x2 = 0

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-48' to each side of the equation.

48 + -48 + 3x2 = 0 + -48

Combine like terms: 48 + -48 = 0

0 + 3x2 = 0 + -48

3x2 = 0 + -48

Combine like terms: 0 + -48 = -48

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Divide each side by '3'.

x2 = -16

Simplifying

x2 = -16

Reorder the terms:

16 + x2 = -16 + 16

Combine like terms: -16 + 16 = 0

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The solution to this equation could not be determined.

8 0
3 years ago
I’ve Tried solving this and i just can’t get it , 64 = 4 + 12g
Alika [10]

Answer:

5

Step-by-step explanation:

64 = 4 + 12g

64 - 4 = 12g

60 = 12g

60 / 12 = g

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8 0
3 years ago
Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

Step-by-step explanation:   We are given to solve the following differential equation :

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

and

y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.

From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

7 0
3 years ago
Hey i need help! ty!​
allsm [11]

Answer:

-3 1/5 < - 3 7/10

Step-by-step explanation:

6 0
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