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1 year ago

I WILL GIVE BRAINLIEST

Mathematics

1 answer:

ryzh [129]1 year ago

4 0

Answer:

See below. No image this time.

Step-by-step explanation:

P(2, 8) → P'(-5, 1), which it went 7 units to the left, and 7 units down. Here is the equation: (x - 7, y - 7).

R(-4, -9) → R'(-4, -2), which went 7 units up. Here is the equation: (x, y + 7).

M(10, -3) → M'(3, 4), which went 7 units left, 7 units up. Here's the equation: (x - 7, y + 7).

K(7, 11) → K'(0, 11), which went 7 units left. Here is the equation: (x - 7, y).

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A square pyramid is shown sitting on its base.

allochka39001 [22]

Answer:

Area of Pyramid = 384 cm²

Step-by-step explanation:

Given

Shape: Square Pyramid

Base: 12 cm

Height: 10 cm

Required

The Surface area.

The surface area of the pyramid is calculated by

1. Calculating the area of the base square

2. Calculating the area of the triangles

3. Adding (1) and (2) above

Having highlighted these points,

Step 1: First we calculate the area of the base square

The dimension of the square is 12cm x 12cm

Let L represent the length of the square (L = 12 cm)

So, Area = L * L

Area = 12 cm * 12 cm

Area = 144 cm²

Step 2: There are four triangles in the above pyramid (because the base is a square; and a square has four sides)

The dimension of each triangle is

Base: 12 cm

Height: 10 cm

First, we calculate the area of one triangle

Area = 0.5 * base * height

Area = 0.5 * 12 cm * 10 cm

Area = 60 cm²

If the area of 1 triangle is 60 cm², the area of 4 triangles would be

Area = 4 * 60 cm²

Area = 240 cm²

Step 3: Adding (1) and (2) above

Area of Pyramid = Area of base square + Area of triangles

Area of Pyramid = 144 cm² + 240 cm²

Area of Pyramid = 384 cm²

Hence, the surface area of the pyramid is 384 cm²

5 0

2 years ago

Solve for c.<br><br> a(c−b)=d

tatiyna

Answer:

c= ab + d/ a

Step-by-step explanation:

Let's solve for c.

a(c-b)=d

Step 1: Add ab to both sides.

-ab + ac + ab = d+ ab

ac = ab + d

Step 2: Divide both sides by a.

ac / a = ab + d / a

c = ab + d/ a

Answer:

c= ab + d/ a

Hope this helps ☝️☝☝

3 0

3 years ago

Enter the ordered pair for point J.

AfilCa [17]

I think it is (1,1.5)

8 0

2 years ago

Read 2 more answers

interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}$

where

$\boldsymbol{T}}$ is the unit tangent vector.

$\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ is the speed of the object

We need to find $\boldsymbol{r}'(t)$ , we know that $\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}$ so

$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

The unit tangent vector is defined by

$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

We need to find the derivative of unit tangent vector

$\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})$