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anygoal [31]
3 years ago
5

HELP I NEED HELP ASAP

Mathematics
1 answer:
hram777 [196]3 years ago
6 0

Answer:A??

Step-by-step explanation:

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A rectangle with a length of L and a width of W has a diagonal of 10 inches. Express the perimeter P of the rectangle as a funct
KatRina [158]
<h2>Answer:</h2>

The expression which represents the perimeter P of the rectangle as a function of L is:

          Perimeter=2(L+\sqrt{100-L^2})

<h2>Step-by-step explanation:</h2>

The length and width of a rectangle are denoted by L and W respectively.

Also the diagonal of a rectangle is: 10 inches.

We know that the diagonal of a rectangle in terms of L and W are given by:

10=\sqrt{L^2+W^2}

( Since, the diagonal of a rectangle act as a hypotenuse of the right angled triangle and we use the Pythagorean Theorem )

Hence, we have:

10^2=L^2+W^2\\\\i.e.\\\\W^2=100-L^2\\\\W=\pm \sqrt{100-L^2}

But we know that width can't be negative. It has to be greater than 0.

Hence, we have:

W=\sqrt{100-L^2}

Now, we know that the Perimeter of a rectangle is given by:

Perimeter=2(L+W)

Here we have:

Perimeter=2(L+\sqrt{100-L^2})

7 0
3 years ago
A road perpendicular to a highway leads to a farmhouse located d miles away. An automobile traveling on this highway passes thro
pshichka [43]

Answer:

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}

Step-by-step explanation:

A road is perpendicular to a highway leading to a farmhouse d miles away.

An automobile passes through the point of intersection with a constant speed \frac{dx}{dt} = r mph

Let x be the distance of automobile from the point of intersection and distance between the automobile and farmhouse is 'h' miles.

Then by Pythagoras theorem,

h² = d² + x²

By taking derivative on both the sides of the equation,

(2h)\frac{dh}{dt}=(2x)\frac{dx}{dt}

(h)\frac{dh}{dt}=(x)\frac{dx}{dt}

(h)\frac{dh}{dt}=rx

\frac{dh}{dt}=\frac{rx}{h}

When automobile is 30 miles past the intersection,

For x = 30

\frac{dh}{dt}=\frac{30r}{h}

Since h=\sqrt{d^{2}+(30)^{2}}

Therefore,

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+(30)^{2}}}

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}

3 0
3 years ago
Factor and find the zeros of the relation.
Nady [450]

Answer:

Step-by-step explanation:

c.

y= 30x² +44x +16

factor

y= 2 (15x² +22x +8)

find zeros

15x²+22x+8=0

ax²+bx+c=0

x= (-b±√(b²-4ac))/2a

x= -22±√484-480/30 = -22±2/30 = -11±1/15

there are 2 zeros x= -10/15 ; x= -12/15

d.

y= 10x² +60x +80

factor

y= 10(x²+6x+8) = 10( x+2)(x+4)

finding zeros

x+2 =0 so x= -2

x+4 =0 so x= -4

there are 2 zeros x= -2; x= -4

4 0
3 years ago
Write 16.66666666... as a fraction.
adoni [48]
The answer is 16 2/3 as explained above

3 0
3 years ago
The Polygons are similar. Find the value of x. ​
ollegr [7]

Answer:

13

Step-by-step explanation:

8/2=16/4

=4

x-3=2.5*4

x=10+3

x=13

8 0
3 years ago
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