Can someone help me out, I am gonna fail :(

Find the area of the shaded region

nexus9112 [7]

Answer:

40

Step-by-step explanation:

area of shaded region = area of whole figure - area of non shaded region

area of whole figure: The whole figure is a rectangle

The area of a rectangle can simply be calculated by multiplying the width by the length

The given width is 10ft and the given length is 8ft

Hence area of whole figure = 10 * 8 = 80ft²

Area of non shaded region: the non shaded region also creates a rectangle

Like stated previously the area of a rectangle can simply be calculated by multiplying the width by the length

The given width is 5ft and the given length is 8ft

Hence, area of non shaded region = 5 * 8 = 40ft²

Finally we can find the area of the shaded region.

We can easily do this my subtracting the area of the nonshaded region from the area of the whole figure

If we have identified that the area of the whole figure is 80 and the area of the non shaded region is 40

Then, area of shaded region = 80 - 40 = 40ft²

6 0

2 years ago

Which of the following pairs of points are both solutions to the equation 2x-5y=6 ? and and and and

zhannawk [14.2K]

Answer:

Any points on the line y=2/5 x - 6/5. See picture below.

Step-by-step explanation:

Convert the equation to slope intercept form and graph it.

2x - 5y = 6

-5y = 6 - 2x

y = -2/-5 x + 6/-5

y=2/5 x - 6/5.

Locate each of the points listed on the graph. If they are a part of the line, then they are solutions.

3 0

3 years ago

Help please!

bagirrra123 [75]

(d): y = mx+n

m = -2/3 ⇒ y = (-2/3)x +n

A(-4, 6) ∈ d ⇒ 6 = (-2/3)·(-4) +n ⇒ 6 = 8/3 +n ⇒

⇒ n = 6 - 8/3 ⇒ n = 10/3

Now, we have:

y = (-2/3)x +10/3

6 0

3 years ago

What is the equation of the line in standard form? 3x−y=−6 3x + y = 6 x + 6y = 9 x−6y=−9

lisabon 2012 [21]

The standard form: Ax + By = C

3x - y = -6 YES - A = 3, B = -1, C = -6

3x + y = 6 YES - A = 3, B = 1, C = 6

x + 6y = 9 YES - A = 1, B = 6, C = 9

x - 6y = -9 YES - A = 1, B = -6, C = -9

8 0

3 years ago

At what point on the paraboloid y = x2 + z2 is the tangent plane parallel to the plane 3x + 2y + 7z = 2? (if an answer does not

Nikitich [7]

If $f(x, y, z) = c$ represent a family of surfaces for different values of the constant $c$ . The gradient of the function $f$ defined as $\nabla f$ is a vector normal to the surface $f(x, y, z) = c$ .

Given the paraboloid

$y = x^2 + z^2$ .

We can rewrite it as a scalar value function f as follows:

$f(x,y,z)=x^2-y+z^2=0$

The normal to the paraboloid at any point is given by:

$\nabla f= i\frac{\partial}{\partial x}(x^2-y+z^2) - j\frac{\partial}{\partial y}(x^2-y+z^2) + k\frac{\partial}{\partial z}(x^2-y+z^2) \\ \\ =2xi-j+2zk$

Also, the normal to the given plane $3x + 2y + 7z = 2$ is given by:

$3i+2j+7k$

Equating the two normal vectors, we have:

$2x=3\Rightarrow x= \frac{3}{2} \\ \\ -1=2 \\ \\ 2z=7\Rightarrow z= \frac{7}{2}$

Since, -1 = 2 is not possible, therefore there exist no such point on the paraboloid $y = x^2 + z^2$ such that the tangent plane is parallel to the plane 3x + 2y + 7z = 2.

4 0

3 years ago