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A rectangle has a width of 46 centimeters and a perimeter of 208 centimeters. What is the rectangles length

alexdok [17]

208- ( 2* 46)= X
208- 92=2x
x=58

The length is 58cm.

Check by:
58*2 +46*2
116+92= 208

4 0

3 years ago

Write an equation of the line with the given slope that contains the given point.

Alchen [17]

Answer:

$y=-2x+12$

Step-by-step explanation:

We can use the point-slope form given by:

$y-y_1=m(x-x_1)$

Where m is the slope and (x₁, y₁) is a point.

So, we will substitute -2 for m and (5, 2) for (x₁, y₁). This gives us:

$y-(2)=-2(x-(5))$

Simplify:

$y-2=-2(x-5)$

Distribute:

$y-2=-2x+10$

Add 2 to both sides. Hence, our equation is:

$y=-2x+12$

3 0

3 years ago

The function f is continuous on the closed interval [1,15] and has the values shown on the table above. Let g(x) = ∫f(t) dt [1,x

Anna11 [10]

We're looking for the two values being subtracted here. One of these values is easy to find:

g(1) = ∫f(t)dt = 0
since taking the integral over an interval of length 0 is 0.

The other value we find by taking a Left Riemann Sum, which means that we divide the interval [1,15] into the intervals listed above and find the area of rectangles over those regions:

Each integral breaks down like so:

(3-1)*f(1)=4

(6-3)*f(3)=9

(10-6)*f(6)=16

(15-10)*f(10)=10.

So, the sum of all these integrals is 39, which means g(15)=39.

Then, g(15)-g(1)=39-0=39.

I hope my answer has come to your help. God bless and have a nice day ahead!

3 0

3 years ago

Read 2 more answers

A disk with radius 3 units is inscribed in a regular hexagon. Find the approximate area of the inscribed disk using the regular

Oksi-84 [34.3K]

Answer:

The approximate are of the inscribed disk using the regular hexagon is $A=18\sqrt{3}\ units^2$