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3 years ago

Which expression correctly represents the perimeter of a rectangle that is 12 centimeters long and 7 centimeters wide?

Mathematics

1 answer:

Goryan [66]3 years ago

3 0

Answer:

The expression for the perimeter of the rectangle is:

P = 2*(12 cm) + 2*(7 cm)

And the perimeter is:

P = 38cm

Step-by-step explanation:

We know that for a rectangle of length L and width W, the perimeter is given by:

P = 2*L + 2*W

So, if we have a rectangle that is 12 cm long, we have L = 12cm

and if the rectangle is 7cm wide, we have W = 7cm

Now we can replace these in the perimeter equation to get:

P = 2*(12 cm) + 2*(7 cm)

P = 24cm + 14cm

P = 38cm

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Let f(x)=x2−2x−4 . What is the average rate of change for the quadratic function from x=−1 to x = 4?

goblinko [34]

Answer:

The average rate of change for f(x) from x=−1 to x = 4 is, 1

Step-by-step explanation:

Average rate A(x) of change for a function f(x) over [a, b] is given by:

$A(x) = \frac{f(b)-f(a)}{b-a}$

As per the statement:

$f(x) = x^2-2x-4$

we have to find the average rate of change from x = -1 to x = 4

At x = -1

$f(-1) = (-1)^2-2(-1)-4 = 1+2-4 = -1$

and

at x = 4

$f(4) = (4)^2-2(4)-4 = 16-8-4 = 4$

Substitute these in [1] we have;

$A(x) = \frac{f(4)-f(-1)}{4-(-1)}$

⇒ $A(x) = \frac{4-(-1)}{4+1}$

⇒ $A(x) = \frac{5}{5}$

Simplify:

A(x) = 1

Therefore, the average rate of change for f(x) from x=−1 to x = 4 is, 1

4 0

3 years ago

Read 2 more answers

Write the given differential equation in the form L(y) = g(x), where L is a linear differential operator with constant coefficie

melamori03 [73]

Answer:

The complete solution is

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

Step-by-step explanation:

Given differential equation is

3y"- 8y' - 3y =4

The trial solution is

$y = e^{mx}$

Differentiating with respect to x

$y'= me^{mx}$

Again differentiating with respect to x

$y''= m ^2 e^{mx}$

Putting the value of y, y' and y'' in left side of the differential equation

$3m^2e^{mx}-8m e^{mx}- 3e^{mx}=0$

$\Rightarrow 3m^2-8m-3=0$

The auxiliary equation is

$3m^2-8m-3=0$

$\Rightarrow 3m^2 -9m+m-3m=0$

$\Rightarrow 3m(m-3)+1(m-3)=0$

$\Rightarrow (3m+1)(m-3)=0$

$\Rightarrow m = 3, -\frac13$

The complementary function is

$y= Ae^{3x}+Be^{-\frac13 x}$

y''= D², y' = D

The given differential equation is

(3D²-8D-3D)y =4

⇒(3D+1)(D-3)y =4

Since the linear operation is

L(D) ≡ (3D+1)(D-3)

For particular integral

$y_p=\frac 1{(3D+1)(D-3)} .4$

$=4.\frac 1{(3D+1)(D-3)} .e^{0.x}$ [since $e^{0.x}=1$ ]

$=4\frac{1}{(3.0+1)(0-3)}$ [ replace D by 0 , since L(0)≠0]

$=-\frac43$

The complete solution is

y= C.F+P.I

$\therefore y= Ae^{3x}+Be^{-\frac13 x}-\frac43$

4 0

3 years ago

4k + 15 greater than -2k + 3

ioda

Answer:

k>-2

Step-by-step explanation:

4k+15>-2k+3

4k-(-2k)+15>3

4k+2k+15>3

6k+15>3

6k>3-15

6k>-12

k>-12/6

k>-2

8 0

3 years ago

A water tank is 3/4 full.

guapka [62]

x/.75=x-56/0.4

0.4x=.75x-42

-0.35x=-42

x=120

I set up a proportion in which x=the tank when it is 3/4 full. When the tank is 3/4 full, solving the proportion tells us the x=120 litres. 120/3=40, and 120+40 is 160, showing that a tank completely full would indeed hold 160 litres.

Hope this helps!

3 0

3 years ago

Give me some tips for the Shsat that is tomorrow.

Aliun [14]

Answer:

1. Sleep well

2. Eat a proper meal to increase your energy

3. Do the ones you are good at first. Skip the harder questions

4. There are 2 parts. One is ELA while the other is math. Do the one that you are good at first so you wouldn't be annoyed.

5. When you are finished filling out the test slip double-check your work that way you could see which one you got wrong.

6. Have faith in yourself. :)

Step-by-step explanation:

8 0

2 years ago