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motikmotik
4 years ago
11

Find the perimeter of rectangle whose length is 40 and the diagonal is 41 cm​

Mathematics
1 answer:
Bond [772]4 years ago
4 0

Answer:

<h2>              P = 98 cm</h2>

Step-by-step explanation:

Given one side and diagonal of rectangle we can use Pythagorean theorem to calculate the other side of it.

L=40\, cm\\D=41\,cm\\W=\ ?\\\\W^2+L^2=D^2\\\\W^2+40^2=41^2\\\\W^2+1600=1681\\\\W^2=81\\\\W=9

Perimeter:

P = 2W + 2L = 2•40 + 2•9 = 80 + 18 = 98

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What is the equation for a slope -1 through the point (2,0)?
RUDIKE [14]

Answer:

y = - x + 2

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = - 1, thus

y = - x + c ← is the partial equation

To find c substitute (2, 0) into the partial equation

0 = - 2 + c ⇒ c = 0 + 2 = 2

y = - x + 2 ← equation of line

4 0
3 years ago
A STORY OF UNITS
Aleks04 [339]

Answer:

<em>$40 </em>will be the total cost of the ribbon.

Step-by-step explanation:

To find the cost of ribbon to be put on the edges of rectangular banner, we need to:

1. Find the perimeter of the rectangle.

2. Multiply the perimeter with the per foot cost to find the total cost.

Let us find the perimeter first:

Perimeter of a rectangle is given as:

P = 2 \times \text{(length+width)}

Here, the dimensions of rectangle are 4ft by 6ft.

So, perimeter = 2 \times (4+6) \Rightarrow 20\ ft

It is given that per foot cost is $2.

Total cost of the ribbon = 2 \times 20 \Rightarrow $40

<em>$40 </em>will be the total cost of the ribbon.

7 0
4 years ago
A lines slope is 15 and its y-intercept is 2. what is its equation in slope intersept form
Amanda [17]

Answer:

y = 15x + 2

Step-by-step explanation:

Use the slope-intercept form:

y = mx + b becomes

y = 15x + 2

5 0
3 years ago
Read 2 more answers
The thickness X of aluminum sheets is distributed according to the probability density function f(x) = 450 (x2 - x) if 6 &lt; x
grandymaker [24]

Solution :

Given :

f(x) = \left\{\begin{matrix}\frac{1}{450}(x^2-x) & \text{if  } 6 < x < 12 \\ 0 & \text{otherwise}\end{matrix}\right.

1. Cumulative distribution function

$P(X \leq x) = \int_{- \infty}^x f(x) \ dx$

              $=\int_{- \infty}^6 f(x) dx + \int_{6}^x f(x) dx $

             $=0+\int_6^x \frac{1}{450}(x^2-x) \ dx$

             $=\frac{1}{450} \int_6^x (x^2-x) \ dx$

             $=\frac{1}{450}\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_6^x$

             $=\frac{1}{450}\left[ \left( \frac{x^3}{3} - \frac{x^2}{2}\left) - \left( \frac{6^3}{3} - \frac{6^2}{2} \right) \right]  $

            $=\frac{1}{450}\left[\frac{x^3}{3} - \frac{x^2}{2} - 54 \right]$

2.  Mean $E(x) = \int_{- \infty}^{\infty} \ x \ f(x) \ dx$

                       $=\int_{6}^{12}x . \left( \frac{1}{450} \ (x^2-x)\right)\  dx$

                     $=\frac{1}{450} \int_6^{12} \ (x^3 - x^2) \ dx$

                     $=\frac{1}{450} \left[\frac{x^4}{4} - \frac{x^3}{3} \right]_6^{12} \ dx$

                     $=\frac{1}{450} \left[ \left(\frac{(12)^4}{4} - \frac{(12)^3}{3} \right) -  \left(\frac{(6)^4}{4} - \frac{(6)^3}{3} \right) $

                     $=\frac{1}{450} [4608 - 252]$

                    = 17.2857

5 0
3 years ago
Round the number to the place value given. 1,370,945 rounded to the ten-thousand is.
Monica [59]

Explanation:

The question requires that we round the given number to Ten-thousand

To do so, we will have to write the place values of the numbers first

Next, we find the digits before the Ten-thousand digits and then approximate the value

One of the principles to use is that

numbers from 0 to 4 are rounded to 0

while numbers from 5 to 9 are rounded to 1

In our case, The thousand digit is 0, so we round down to 0, then we add to the Ten-thousand digits and convert each of the numbers before the Ten-thousand digits to 0

Thus, we will have

1,370,000

Hence, the answer is 1,370,000

7 0
1 year ago
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