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3 years ago

Geometric mean of 45 and 15

Mathematics

2 answers:

amm18123 years ago

4 0

Answer:

Step-by-step explanation:

$\frac{45+15}{2}$ equals 30.

posledela3 years ago

3 0

Answer:

45 + 15 = 60 divided by 2 is 30

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Equilateral triangle ABC has a perimeter of 96 millimeters. A perpendicular bisector is drawn from angle A to side BC at point M

a_sh-v [17]

Answer:

16 mm

Step-by-step explanation:

Since ABC is an equilateral triangle, all three sides are the same length. The perimeter is found by adding together all of these equal sides; letting x represent the length of a side of the triangle, this gives us

x + x + x = 96

Combining like terms,

3x = 96

Dividing both sides by 3,

3x/3 = 96/3

x = 32

Since AM is a perpendicular bisector, it splits BC into two congruent sections. This means the length of MC, half of BC, will be 32/2 = 16.

3 0

3 years ago

Read 2 more answers

A furniture store is having a sale on sofas and you're going to buy one. The advertisers know that buyers get to the store and t

MrRissso [65]

Answer:

$290

Step-by-step explanation:

We are told that 1 out of 5 buyers change to a more expensive sofa than the one in the sale advertisement.

Now we are told that the advertised sofa is $250 and the more expensive sofa is $450.

Thus;

P(x) for expensive sofa = 1/5

P(x) for sofa in sale advertisement = 4/5

Thus, expected value is;

E(X) = (1/5)450 + (4/5)250

E(x) = 90 + 200

E(x) = $290

6 0

3 years ago

What plane represents the top or the box?

natulia [17]

Answer:

it represents the top and the bptto

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4 years ago

The diagram shows how cos θ, sin θ, and tan θ relate to the unit circle. Copy the diagram and show how sec θ, csc θ, and cot θ r

DIA [1.3K]

Copy the diagram and show how sec θ, csc θ, and cot θ relate to the unit circle.

The representation of the diagram is shown if Figure 1. There's a relationship between sec θ, csc θ, and cot θ related the unit circle. Lines green, blue and pink show the relationship.

a.1 First, find in the diagram a segment whose length is sec θ.

The segment whose length is sec θ is shown in Figure 2, this length is the segment $\overline{OF}$ , that is, the line in green.

a.2 Explain why its length is sec θ.

We know these relationships:

(1) $sin \theta=\frac{\overline{BD}}{\overline{OB}}=\frac{\overline{BD}}{r}=\frac{\overline{BD}}{1}=\overline{BD}$

(2) $cos \theta=\frac{\overline{OD}}{\overline{OB}}=\frac{\overline{OD}}{r}=\frac{\overline{OD}}{1}=\overline{OD}$

(3) $tan \theta=\frac{\overline{FD}}{\overline{OC}}=\frac{\overline{FC}}{r}=\frac{\overline{FC}}{1}=\overline{FC}$

Triangles ΔOFC and ΔOBD are similar, so it is true that:

$\frac{\overline{FC}}{\overline{OF}}= \frac{\overline{BD}}{\overline{OB}}$ 

∴ $\overline{OF}= \frac{\overline{FC}}{\overline{BD}}= \frac{tan \theta}{sin \theta}= \frac{1}{cos \theta} \rightarrow \boxed{sec \theta= \frac{1}{cos \theta}}$ 

b.1 Next, find cot θ

The segment whose length is cot θ is shown in Figure 3, this length is the segment $\overline{AR}$ , that is, the line in pink.

b.2 Use the representation of tangent as a clue for what to show for cotangent.

It's true that:

$\frac{\overline{OS}}{\overline{OC}}= \frac{\overline{SR}}{\overline{FC}}$

But:

$\overline{SR}=\overline{OA}$
$\overline{OS}=\overline{AR}$

Then:

$\overline{AR}= \frac{1}{\overline{FC}}= \frac{1}{tan\theta} \rightarrow \boxed{cot \theta= \frac{1}{tan \theta}}$

b.3 Justify your claim for cot θ.

As shown in Figure 3, θ is an internal angle and ∠A = 90°, therefore ΔOAR is a right angle, so it is true that:

$cot \theta= \frac{\overline{AR}}{\overline{OA}}=\frac{\overline{AR}}{r}=\frac{\overline{AR}}{1} \rightarrow \boxed{cot \theta=\overline{AR}}$

c. find csc θ in your diagram.

The segment whose length is csc θ is shown in Figure 4, this length is the segment $\overline{OR}$ , that is, the line in green.

3 0

4 years ago

A = 4 0 0 1 3 0 −2 3 −1 Find the characteristic polynomial for the matrix A. (Write your answer in terms of λ.) Find the real ei

Illusion [34]

Answer:

Step-by-step explanation:

We are given the matrix

$A = \left[\begin{matrix}4&0&0 \\ 1&3&0 \\-2&3&-1 \end{matrix}\right]$

a) To find the characteristic polynomial we calculate $\text{det}(A-\lambda I)=0$ where I is the identity matrix of appropiate size. in this case the characteristic polynomial is

$\left|\begin{matrix}4-\lambda&0&0 \\ 1&3-\lambda&0 \\-2&3&-1-\lambda \end{matrix}\right|=0$

Since this matrix is upper triangular, its determinant is the multiplication of the diagonal entries, that is

$(4-\lambda)(3-\lambda)(-1-\lambda)=(\lambda-4)(\lambda-3)(\lambda+1)=0$

which is the characteristic polynomial of A.

b) To find the eigenvalues of A, we find the roots of the characteristic polynomials. In this case they are $\lambda=4,3,-1$

c) To find the base associated to the eigenvalue lambda, we replace the value of lambda in the expression $A-\lambda I$ and solve the system $(A-\lambda I)x =0$ by finding a base for its solution space. We will show this process for one value of lambda and give the solution for the other cases.

Consider $\lambda = 4$ . We get the matrix

$\left[\begin{matrix}0&0&0 \\ 1&-1&0 \\-2&3&-5 \end{matrix}\right]$

The second line gives us the equation x-y =0. Which implies that x=y. The third line gives us the equation -2x+3y-5z=0. Since x=y, it becomes y-5z =0. This implies that y = 5z. So, combining this equations, the solution of the homogeneus system is given by

$(x,y,z) = (5z,5z,z) = z(5,5,1)$

So, the base for this eigenspace is the vector (5,5,1).

If $\lambda = 3$ then the base is (0,4,3) and if $\lambda = -1$ then the base is (0,0,1)

3 0

3 years ago

Geometric mean of 45 and 15​

Geometric mean of 45 and 15