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3 years ago

PLEASE HELP!!! WILL MARK BRAINLEIST

Mathematics

1 answer:

Ksju [112]3 years ago

6 0

The volume is 1494.35
The formula is length times height adding the bases than divided it by 2

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The decimal expression for 8 km 5 m is

Neko [114]

Answer:

8.005 km

Step-by-step explanation:

There are 1000 meters in a kilometer, so:

5/1000 = 1/200 = 0.005

8 km + 0.005 km = 8.005 km

3 0

4 years ago

Rewrite 4 x 4 x 4 x 4 x4 using a exponent

alukav5142 [94]

Answer:4 to the 4th power

Step-by-step explanation:

4^4

3 0

3 years ago

Read 2 more answers

Evaluate the discriminant.<br> Exact answer (no rounding)<br> 0= -4x – 24x – 38

frutty [35]

Answer: x= -19/14 —> (19 over 14)

8 0

3 years ago

Serhud [2]

A circle is characterized by radius, arc, sectors and circumference

The length of the major arc is $18.75\pi$
The radius of the circle is 15
The area of the shaded sector is $140.625 * \pi$

<h3>Length of the major arc</h3>

The given parameters are:

$C = 30\pi$ --- the circumference

$\theta = 225^o$ -- the center angle

The length of the major arc is calculated using:

$L = \frac{\theta}{360} * C$

So, we have:

$L = \frac{225}{360} *30\pi$

Evaluate

$L = 18.75\pi$

Hence, the length of the major arc is $18.75\pi$

<h3>The radius of the circle</h3>

The circumference is given as:

$C = 30\pi$

So, we have:

$2\pi r = 30\pi$

Divide through by 2pi

$r = 15$

Hence, the radius of the circle is 15

<h3>The area of the shaded sector</h3>

The area of a sector is:

$A = \frac{\theta}{360} * \pi r^2$

So, we have:

$A = \frac{225}{360} * \pi * 15^2$

Evaluate

$A = 140.625 * \pi$

Hence, the area of the shaded sector is $140.625 * \pi$

Read more about circumference at:

brainly.com/question/15673093

6 0

2 years ago

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

4 years ago