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2 years ago

A rectangle. one side is 2x + 7 and the other is x+3

Mathematics

1 answer:

saveliy_v [14]2 years ago

6 0

$\large{\green}\fcolorbox{purple}{pink}{\bf{\underline{\red{\color{red}Answer}}}}$

6x + 20

Step-by-step explanation:

To find :-

Perimeter of rectangle

Given :-

Length = 2x + 7

Breadth/width = x + 3

Solution :-

We know that

$\mathfrak {perimeter \: of \: rectangle =2(length + breadth) }$

Now we will substitute the values of length and breadth

$\sf \longmapsto perimeter = 2 \{(2x + 7) + (x + 3) \} \\ \\ \\ \sf \longmapsto perimeter = 2 \{3x + 10 \} \\ \\ \\ \sf \longmapsto perimeter = 6x + 20$

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Simplify the following problem. (-8+61) +(6+61)

Travka [436]

Answer: -8+61 = 53 and 6+61 = 67

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3 years ago

Alan and Margot each drive from City A to City B, a distance of 147 miles. They take the same route and drive at constant speeds

Westkost [7]

Answer:

Margot gets from A to B faster

Step-by-step explanation:

Alan takes 2 hr and 35 min = 2. 58... hr

Margot takes 147/64 hr = 2.296.. hr

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3 years ago

Please answer as soon as possible PLEASE

tatuchka [14]

Answer:

x>-5

Step-by-step explanation:

-5(x+7)<-10

divide by -5 (and because we divided by a negative number swap the inequality sign)

x+7>2

x> -5

7 0

2 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

3 years ago

Show work please I need help asap

rosijanka [135]

Answer:

12. -11

13. 2

Step-by-step explanation:

12. First recall the order of operations (BEDMAS): Brackets, Exponents, Division, Multiplication, Addition, Subtraction

For question 12 there is a division and its priority is before other operations in the equation so you must divide 16/-2 first which gives you -8.

From there because addition and subtraction is on the same level of order, you would do the question straight as it is shown (but replace 16/-2 with -8.

Your new equation is -10-8+7 = -11

13. follow the same as above using BEDMAS. It may help to look at it with additional brackets: ((-68)/(-4)) + ((5 x (-3))

keep in mind there’s an addition between the two

Division and multiplication is on the same level in BEDMAS so first divide -68/-4 = 17

Second multiply 5 x -3 = -15

Now you can combine the two answers using addition: 17 + (-15) = 2

5 0

3 years ago