A 96cm A long wire is given the shape of a rectangle such that its length is 12 cm more than its breadth Find the dimensions of the rectangle.​

Answer:

[tex]\begin{gathered} \: \end{gathered}[/tex]

[tex]\begin{gathered} \: \end{gathered}[/tex]

Answer:151 in.³

Step-by-step explanation:

Answer:

T = 31

Step-by-step explanation:

(T + 8) + T = 70

subtract 8 from both sides

T+T = 62

Rewrite

2T = 62

divide both sides by 2

T = 31

Answer:

i'm not sure but here's some encouragement- YOU CAN DO IT

Step-by-step explanation:

heres a link hope it help you

https://www.journaldespalaces.com/carriere/offre-118597-Officier-de-restauration-Cafetier.html

Answer:

Step-by-step explanation:

The factored form of a quadratic function is f (x) = a (x - p) (x - q) where p and q are the zeros of f (x).

Simplify the integrand as

[tex]\dfrac{2x^3 - x^2 - 2x + 4}{1 + x^2} = \dfrac{(2x^3 + 2x) - (x^2 + 1) - 4x + 5}{x^2 + 1} \\\\ = \dfrac{2x(x^2 + 1) - (x^2 + 1) - 4x + 5}{x^2 + 1} \\\\ = 2x - 1 - \dfrac{4x - 5}{x^2 + 1}[/tex]

(in other words, compute the quotient and remainder)

We can further split up and prepare the remainder term for integration by rewriting it as

[tex]\dfrac{4x - 5}{x^2 + 1} = 2\times\dfrac{2x}{x^2 + 1} - \dfrac5{x^2 + 1}[/tex]

Now we integrate:

[tex]\displaystyle \frac{2x^3 - x^2 - 2x + 4}{1 + x^2} \, dx = \int \left(2x - 1 - 2\times\frac{2x}{x^2+1} + \frac5{x^2+1}\right) \, dx[/tex]

[tex]\displaystyle = x^2 - x - 2 \int \frac{2x}{x^2+1} \, dx + 5 \int \frac{dx}{x^2+1}[/tex]

In the first remaining integral, substitute y = x² + 1 and dy = 2x dx. In the last integral, recall that d/dx [arctan(x)] = 1/(x² + 1).

[tex]\displaystyle = x^2 - x - 2 \int \frac{dy}y + 5 \int \frac{dx}{x^2+1}[/tex]

[tex]\displaystyle = x^2 - x - 2 \ln|y| + 5 \arctan(x) + C[/tex]

[tex]\displaystyle = \boxed{x^2 - x - 2 \ln(x^2+1) + 5 \arctan(x) + C}[/tex]

Answer:

you would have to separate the shapes and find out what the area  of them both and add them together

Step-by-step explanation:

Answer:

$9,360

Explanation:

so for 1040 hours, he earned:

$9 * 1,040

$9,360

Translation involves moving a function along its coordinates

The image of the function is [tex]g(x) = (x + 2)^2 - 6[/tex]

The parent function is given as:

[tex]f(x) = x^2[/tex]

When the function is translated 2 units left, we have:

[tex]f'(x) = (x + 2)^2[/tex]

When the function is translated 6 units down, we have:

[tex]f"(x) = (x + 2)^2 - 6[/tex]

Rewrite as:

[tex]g(x) = (x + 2)^2 - 6[/tex]

Hence, the image of the function is [tex]g(x) = (x + 2)^2 - 6[/tex]

Read more about translation at:

https://brainly.com/question/11468584

Given :

⠀

To Find :

⠀

Solution :

We know that,

[tex]\qquad { \pmb{ \bf{Length \times Width = Area_{(rectangle)}}}}\:[/tex]

⠀

Now substituting the values :

[tex]\qquad {\dashrightarrow{ \sf{214 \: in\times 234 \: in = Area_{(rectangle)}}}}\:[/tex]

[tex]\qquad {\dashrightarrow \: { { \sf{50,076 \: {in}^{2} = Area_{(rectangle)}}}}\:}

[/tex]

⠀

Hence, The area of the rectangle is 50,076 in² .

The area of a rectangle with a length of 2 1/4 inches and a width of 2 3/4 inches is 6 3/16 in². Option C is correct.

Area of a rectangle is the product of the length of the rectangle and the width of the rectangle. It can be given as,

[tex]A=a\times b[/tex]

Here, (a)is the length of the rectangle and (b) is the width of the rectangle.

The rectangle given in the problem has,

Thus, the area of the rectangle is,

[tex]A=2\dfrac{1}{4}\times2\dfrac{3}{4}\\A=\dfrac{8+1}{4}\times\dfrac{8+3}{4}\\A=\dfrac{9}{4}\times\dfrac{11}{4}\\A=\dfrac{99}{16}\\\A=\dfrac{96+3}{16}\\A=6\dfrac{3}{16}[/tex]

Thus, the area of a rectangle with a length of 2 1/4 inches and a width of 2 3/4 inches is 6 3/16 in². Option C is correct.

Learn more about the area of rectangle here;

https://brainly.com/question/11202023

Answer:

where is the table?

Step-by-step explanation:

._.

Answer:

The answer is B 80.00 hope this helps

have a nice day

Step-by-step explanation:

Answer:

"Through" (and any subsequent words) was ignored because we limit queries to 32 words.

The required equation that can be used to figure out how many miles Hugh and his brothers have traveled based on the number of miles they have driven is y = 65x

Given the following variables

Time taken to travel is x

Mils traveled is y

Using the coordinate points (2, 130) and (4, 260)

The standard linear equation is y = mx + b

slope m = 260-130/4-2
m = 130/2

m = 65

For the intercept

130 = 65(2) + b

b = 0

The required equation that can be used to figure out how many miles Hugh and his brothers have traveled based on the number of miles they have driven is y = 65x

Learn more on linear functions here; https://brainly.com/question/14323743

We need slope

[tex]\\ \rm\Rrightarrow m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

[tex]\\ \rm\Rrightarrow m=\dfrac{2-1}{6-3}[/tex]

[tex]\\ \rm\Rrightarrow m=\dfrac{1}{3}[/tex]

This can be solved using two methods:

Method #1:

The slope is 1/3.

Method #2:

The slope is 1/3.

Looking at the two methods, we can conclude that:

The slope of the line is 1/3.

Answer:

If he earns (e) $7.75 per hour, this implies e = 7.75h.

Step-by-step explanation:

If you wanted to solve that-

e = 7.75h

193.75 = 7.75h

Divide both sides by 7.25  

h = 25

Answer:

Step-by-step explanation:

1 3 5 are all odd numbers.

There are 6 numbers all together on a die. 1 2 3 4 5 6

That means that 3/6 times, he should roll an odd number

3/6 * 78 = 39

He should get 39 rolls which are an odd number.

Answer:

300

Step-by-step explanation:

60 * 5 = 300

Answer:

Point P is the center of Rotation.

Answer:

[tex]\huge\boxed{\sf{b=55\:(b=y-intercept\:;)}}[/tex]

Step-by-step explanation:

Hello.

The y-intercept of a line is a point where the graph touches the y-axis.

Y-intercepts always have an x-coordinate of 0.

Now, what is the y-intercept?

If we have a point (0, b) then b is the y-intercept.

This is just a formula, so don't think that the y-intercept is b, okay?

The y-intercept is b, but only in formulae. :)

Using the formula, we can deduce that

the y-intercept is 55.

I hope it helps.

Have a great day.

[tex]\boxed{imperturbability}[/tex]

Answer:

The y-intercept is the point at which the graph crosses the y-axis. At this point, the x-coordinate is zero. To determine the x-intercept, we set y equal to zero and solve for x. Similarly, to determine the y-intercept, we set x equal to zero and solve for y.

Step-by-step explanation:

hope it helps

Answer:

above is the solution to the question

Answer:

Step-by-step explanation:

Expression => | -2 |

Equation => 6 - 1 = 2 + 3

Inequality => 8^m > 64

Answers:

Expression is [tex]|-2|[/tex]

Equation is [tex]6-1 = 2+3[/tex]

Inequality is [tex]8^m > 64[/tex]

====================================================

Explanation:

If it has an equal sign, then it's an equation.

If it has an inequality sign of one of the following [tex]< , \ > , \ \le, \ \ge[/tex], then it's an inequality.

Otherwise, it's an expression.

Decompose each given F(s) into partial fractions.

[tex]F(s) = \dfrac{s+1}{s(s-1)(s-3)}[/tex]

has partial fraction decomposition

[tex]\dfrac{s+1}{s(s-1)(s-3)} = \dfrac as + \dfrac b{s-1} + \dfrac c{s-3}[/tex]

Combine the rational terms on the right and solve for the coefficients:

[tex]\dfrac{s+1}{s(s-1)(s-3)} = \dfrac{a(s-1)(s-3) + b s(s-3) + c s(s-1)}{s (s-1) (s-3)}[/tex]

[tex]1 = a(s-1)(s-3) + bs(s-3) + c s(s-1)[/tex]

[tex]1 = 3 a + (-4 a - 3 b - c) s + (a + b + c) s^2[/tex]

[tex]\begin{cases}3a=1 \\ -4a-3b-c = 0 \\ a+b+c=0 \end{cases} \implies a=\dfrac13, b=-\dfrac12, c=\dfrac16[/tex]

Then

[tex]F(s) = \dfrac13 \times \dfrac1s - \dfrac12 \times \dfrac1{s-1} + \dfrac16 \times \dfrac1{s-3}[/tex]

Using the frequency-shifting property, the inverse transform is

[tex]\boxed{f(t) = \dfrac13 - \dfrac{e^t}2 + \dfrac{e^{3t}}6}[/tex]

The other transform can be dealt with in the same manner.

[tex]F(s) = \dfrac1{(s-1)(s-2)(s-3)} = \dfrac a{s-1} + \dfrac b{s-2} + \dfrac c{s-3}[/tex]

[tex]\implies 1 = a(s-2)(s-3) + b(s-1)(s-3) + c(s-1)(s-2)[/tex]

[tex]\implies 1 = 6 a + 3 b + 2 c + (-5 a - 4 b - 3 c) s + (a + b + c) s^2[/tex]

[tex]\implies \begin{cases}6 a + 3 b + 2 c=1 \\ -5a-4b-3c = 0 \\ a+b+c=0\end{cases} \implies a=\dfrac12, b=-1, c=\dfrac12[/tex]

[tex]\implies F(s) = \dfrac12 \times \dfrac1{s-1} - \dfrac1{s-2} + \dfrac12 \times \dfrac1{s-3}[/tex]

[tex]\implies \boxed{f(t) = \dfrac{e^t}2 - e^{2t} + \dfrac{e^{3t}}2}[/tex]

[tex]f(x)=cos(x)x^{-2}\implies f(x)=\cfrac{cos(x)}{x^2}\implies \cfrac{df}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{-sin(x)\cdot x^2-cos(x)\cdot 2x}{(x^2)^2}} \\\\\\ \cfrac{df}{dx}=\cfrac{-x[x\cdot sin(x)+2cos(x)]}{x^4}\implies \cfrac{df}{dx}=\cfrac{-[xsin(x)+2cos(x)]}{x^3}[/tex]