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xxTIMURxx [149]

2 years ago

11

A porch has the shape of a quarter circle with a diameter of 5 ft. What is the approximate area of the porch, rounded to the nea

rest square foot? Use 3.14 to approximate pi.
The answer I got is 4.9ft

1 answer:

miss Akunina [59]2 years ago

6 0

Area of a quarter circle = ( pi x r^2 ) / 4
(3.14 x 2.5^2) /4 = 4.9 feet

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All vectors are in Rn. Check the true statements below:

Oduvanchick [21]

Answer:

A), B) and D) are true

Step-by-step explanation:

A) We can prove it as follows:

$Proy_{cv}y=\frac{(y\cdot cv)}{||cv||^2}cv=\frac{c(y\cdot v)}{c^2||v||^2}cv=\frac{(y\cdot v)}{||v||^2}v=Proy_{v}y$

B) When you compute the product Ax, the i-th component is the matrix of the i-th column of A with x, denote this by Ai x. Then, we have that $||Ax||=\sqrt{(A_1 x)^2+\cdots (A_n x)^2}$ . Now, the colums of A are orthonormal so we have that (Ai x)^2=x_i^2. Then $||Ax||=\sqrt{(x_1)^2+\cdots (x_n)^2}=||x||$ .

C) Consider $S=\{(0,2),(2,0)\}\subseteq \mathbb{R}^2$ . This set is orthogonal because $(0,2)\cdot(2,0)=0(2)+2(0)=0$ , but S is not orthonormal because the norm of (0,2) is 2≠1.

D) Let A be an orthogonal matrix in $\mathbb{R}^n$ . Then the columns of A form an orthonormal set. We have that $A^{-1}=A^t$ . To see this, note than the component $b_{ij}$ of the product $A^t A$ is the dot product of the i-th row of $A^t$ and the jth row of $A$ . But the i-th row of $A^t$ is equal to the i-th column of $A$ . If i≠j, this product is equal to 0 (orthogonality) and if i=j this product is equal to 1 (the columns are unit vectors), then $A^t A=I$

E) Consider S={e_1,0}. S is orthogonal but is not linearly independent, because 0∈S.

In fact, every orthogonal set in R^n without zero vectors is linearly independent. Take a orthogonal set $\{u_1,u_2\cdots u_p\}$ and suppose that there are coefficients a_i such that $a_1u_1+a_2u_2\cdots a_nu_n=0$ . For any i, take the dot product with u_i in both sides of the equation. All product are zero except u_i·u_i=||u_i||. Then $a_i||u_i||=0$ then $a_i=0$ .

5 0

3 years ago

Find the equation of a circle with a center at (–7, –1) where a point on the circle is (–4, 3).

gogolik [260]

Answer:

(x + 7)^2 + (y + 1)^2 = 25.

Step-by-step explanation:

The center of a circle is easy to set up. According to the formula below, the formula for the circle will be (x - a)^2 + (y - b)^2 = r^2.

In this case, a = -7 and b = -1, so we have...

(x - (-7))^2 + (y - (-1))^2 = r^2

(x + 7)^2 + (y + 1)^2 = r^2

To get the radius, we need to find the distance between the center and the point on the circle. The distance formula is d = sqrt((x2 - x1)^2 + (y2 - y1)^2).

In this case, x2 = -4, x1 = -7, y2 = 3, and y1 = -1.

sqrt((-4 - -7)^2 + (3 - -1)^2) = sqrt((-4 + 7)^2 + (3 + 1)^2) = sqrt((3)^2 + (4)^2) = sqrt(9 + 16) = sqrt(25) = plus or minus 5.

Since distance can only be positive, the distance is 5 units, meaning that the radius is 5 units.

5^2 = 25

So, your equation should be (x + 7)^2 + (y + 1)^2 = 25.

Hope this helps!

8 0

3 years ago

Someone please help me!!

eduard

On what do you need help

4 0

3 years ago

What is the product?<br><br> (y3+3y+7) x (8y2+y+1)

Nataly [62]

Answer:

I believe that this is what you mean. (y^3+3y+7)*(8y^2+y+1)

Step-by-step explanation:

So your answer would be: 8y^5+y^4+25y^3+59y^2+10y+7

Have a nice day!

I really hope this helps!

5 0

2 years ago

kifflom [539]

Answer:

2

Step-by-step explanation:

7 0

2 years ago