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Natasha_Volkova [10]
4 years ago
12

The length of a rectangle is increasing at a rate of 6 cm/s and the width is increasing at a rate of 4 cm/s. when the length is

16 cm and the width is 12 cm, how fast is the area of the rectangle increasing?
Mathematics
1 answer:
alukav5142 [94]4 years ago
7 0
The area is 192 and the speed it’s increasing at is 24 cm^2/s
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Enter an algebraic equation for the sentence. Use x as your variable.
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Answer:

4x-17=x/2+8

Step-by-step explanation:

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3 years ago
PLEASE HELP!! Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Match the trigonometric rat
Monica [59]

this is the answer on edmentum

6 0
3 years ago
Find the equation of the graphed line.
belka [17]
<h3><u>Answer:</u></h3>

\boxed{\boxed{\pink{\bf \leadsto Hence \ option\ [d]\ \bigg(y =  \dfrac{5}{2}x + 5\bigg) \ is \ correct  }}}

<h3><u>Step-by-step explanation:</u></h3>

Here from the given graph we can see that the graph the graph intersects x axis at (2,0) and y axis at (5,0). On seeing options it's clear that we have to use Slope intercept form . Which is :-

\large\boxed{\orange{\bf y = mx + c} }

We know that slope is \tan\theta. So here slope will be ,

\red{\implies slope = \tan\theta} \\\\\implies slope =\dfrac{PERPENDICULAR}{BASE} \\\\\bf \implies slope =\dfrac{5}{2}

Hence the slope is 5/2 . And here value of c will be 5 since it cuts y axis at (5,0).

\purple{\implies y = mx + c }\\\\\implies y = \dfrac{5}{2}x + 5 \\\\\underline{\boxed{\red{\tt \implies y = \dfrac{5}{2}x + 5 }}}

<h3><u>Hence</u><u> </u><u>option</u><u> </u><u>[</u><u> </u><u>d</u><u> </u><u>]</u><u> </u><u>is</u><u> </u><u>corre</u><u>ct</u><u> </u><u>.</u><u> </u></h3>
4 0
3 years ago
Use the distance formula, show that the points (4,0), (2,1), and (-1,-5) form the vertices of a right triangle
mario62 [17]

Step-by-step explanation:

The distance formula between two points:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Substitute the coordinates of the points.

A(4,\ 0),\ B(2,\ 1),\ C(-1,\ -5)\\\\AB=\sqrt{(2-4)^2+(1-0)^2}=\sqrt{(-2)^2+1^2}=\sqrt{4+1}=\sqrt5\\\\AC=\sqrt{(-1-4)^2+(-5-0)^2}=\sqrt{(-5)^2+(-5)^2}=\sqrt{25+25}=\sqrt{50}\\\\BC=\sqrt{(-1-2)^2+(-5-1)^2}=\sqrt{(-3)^2+(-6)^2}=\sqrt{9+36}=\sqrt{45}

If a ≤ b < c are the sides of the right triangle, then

a² + b² = c²

\sqrt5

used (\sqrt{a})^2=a for a ≥ 0.

AB^2+BC^2=AC^2 therefore ΔABC is a right triangle.

6 0
3 years ago
Please help I'm desperate...
Evgen [1.6K]
Ok here’s the answer! I again had to crop out the first step. here it was to divide equation 1 by 3 and add that to equation 2. all you need do you show that step is to write equation 1: -3x+2y=8 and equation 2: (11y/3)=35/3

6 0
3 years ago
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