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3 years ago

Can you find the perimeter of a rectangle if you only know its area? what about a square? explain.

1 answer:

4 0

I don’t think you can find the perimeter of a rectangle just by knowing the area because a rectangle has different sides (they aren’t the same). But it is possible for a square because in a square, all sides are the same. Let’s say the area of a square is 25^3. To find the perimeter, we just need to find one side than multiply by 4. If we use 25^3 as the area, to find a side we need to find the square root of 25 which is 5 (5x5=25). Now we multiply 5 by 4 which equals 20. 20 will be the perimeter.

Hope this helps and hope you have a great day. Please make brainiest

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Find the inverse of the function.

Akimi4 [234]

Answer:

Step-by-step explanation:

The inverse is found by interchanging the x and y values and solve the result for y.

I will use y for f(x)

y = 4x^2

Inverse

x = 4y^2 Divide by 4

x/4 = 4y^2 /4

x/4 = y^2 Take the square root of both sides.

sqrt(x/4) = sqrt(y^2)

y = sqrt(x/4)

g(x) = sqrt(x/4) If you have a very fussy teacher, the inverse function could be written as

g(x) = +/- sqrt(x)/2

I expect that if the question is computer marked, I would try g(x) = sqrt(x)/2 first.

3 0

3 years ago

F(x) = 9x^2+30x+25 pls help

lapo4ka [179]

Answer: f(x) =(3x+5)^2

x= -5/3

Step-by-step explanation:

5 0

3 years ago

Slope of line that passes through points (2,1) and (17,-17)

mojhsa [17]

Answer: y = -1,2x + 3.4

Step-by-step explanation:

y = ax + b

(2,1) => 1 = 2a + b (1)

(17,-17) => -17 =17a + b (2)

Using caculator Mode => 5 =>1

We get : a=- 1.2 , b = 3.4

=> y = -1.2x + 3.4

7 0

3 years ago

Examine the following system of inequalities.

topjm [15]

<h2>Answer:</h2>

Dotted linear inequality shaded below passes through (0, 4) & (4,3). Dotted parabolic inequality shaded above passes through points (negative 6,4), (negative 4, 0) & (negative 2, 4).

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

Hello! Let me help you to find the correct option to this problem. First of all, we have the following system of inequalities:

$\left\{ \begin{array}{c}y< -\frac{1}{4}x+4\\y>(x+4)^{2}\end{array}\right.$

To solve this, let's write the following equations:

<h3>FIRST:</h3>

$y=-\frac{1}{4}x+4$

This is a linear function written in slope-intercept form as $y=mx+b$ . So, the slope $m=-\frac{1}{4}$ and the y-intercept is $b=4$ . Since in the inequality we have the symbol < then the graph of the line must be dotted. To get the shaded region, let's take a point, say, $(0, 0)$ and let's test whether the region is above or below the graph. So:

$y< -\frac{1}{4}x+4 \\ \\ Let \ x=y=0 \\ \\ 0$

Since the expression is true, then the region is the one including point $(0, 0)$ , that is, it's shaded below.

<h3>SECOND:</h3>

$y=(x+4)^{2}$

This is a parabola that opens upward and whose vertex is $(-4,0)$ . Since in the inequality we have the symbol > then the graph of the parabola must be dotted. Let's take the same point $(0, 0)$ to test whether the region is above or below the graph. So:

$y>(x+4)^{2} \\ \\ Let \ x=y=0 \\ \\ 0>(0+4)^2\\ \\ 0>16 \ False!$

Since the expression is false, then the region is the one that doesn't include point $(0, 0)$ , that is, it's shaded above

____________________

<h3>On the other hand, testing points (0, 4) and (4,3) on the linear function:</h3>

$y=-\frac{1}{4}x+4 \\ \\ \\ \bullet \ (0,4): \\ \\ y=-\frac{1}{4}(0)+4 \therefore y=4 \\ \\ \\ \bullet \ (4,3): \\ \\ y=-\frac{1}{4}(4)+4 \therefore y=3$

So the line passes through these two points.

<h3 /><h3>Now, testing points (negative 6,4), (negative 4, 0) & (negative 2, 4) on the parabola:</h3><h3 />

$y=(x+4)^2 \\ \\ \\ \bullet \ (-6,4): \\ \\ y=(-6+4)^2 \therefore y=(-2)^2 \therefore y=4 \\ \\ \\ \bullet \ (-4,0): \\ \\ y=(-4+4)^2 \therefore y=0 \\ \\ \\ \bullet \ (-2,4): \\ \\ y=(-2+4)^2 \therefore y=(2)^2 \therefore y=4$