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2 years ago

In △ABC, the altitudes from vertices B and C intersect at point M, so that BM = CM. Prove that △ABC is isosceles.

Mathematics

1 answer:

AlekseyPX2 years ago

3 0

Answer:

m∠MBC=m∠MCB by reason base angle theorem

Step-by-step explanation:

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What is 1/7 divided by ? = 14

user100 [1]

First you multiply the numbers you do have to get the number your looking for 1/7•14=2 and 2divided by 1/7=14 so 2 would be your answer

4 0

3 years ago

A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2

andrey2020 [161]

Answer:

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

Step-by-step explanation:

Consider the provided matrix.

$v_1=\begin{bmatrix}-3\\1 \end{bmatrix}$

$v_2=\begin{bmatrix}-1\\1 \end{bmatrix}$

$\lambda_1=4, \lambda_2=2$

The general solution of the equation $x'=Ax$

$x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}$

Substitute the respective values we get:

$x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}$

Substitute initial condition $x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}$

$\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}$

Reduce matrix to reduced row echelon form.

$\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}$

Therefore, $c_1=2.5,c_2=1.5$

Thus, the general solution of the equation $x'=Ax$

$x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

6 0

3 years ago

Consider the system of equations. 20 POINTS NEED HELP ASAP

Alex

- 5x - 1 = 2x + 6

- 7x = 7

7x = -7

x = -1

3 0

2 years ago

A box of chocolates contains 18 chocolate squares, tens of the square are milk chocolate, 5 are dark chocolate, and 3 are.white

Step2247 [10]

Answer:

0.167 ; 0.093

Step-by-step explanation:

Given that:

Total number of chocolate squares = 18 of which÷

Number of milk chocolate square = 10

Dark chocolate square = 5

White chocolate square= 3

Probability of choosing a white chocolate square :

Probability = required outcome / Total possible outcomes

Required outcome = white chocolate square = 3

Number of chocolate square = 18

P(choosing a white chocolate square) = 3 / 18 = 1/6 =

B) probability of randomly selecting a milk chocolate first, replace it, and then select a white chocolate?

With replacement :

First pick :

Number milk chocolate = 10

P(choosing a milk chocolate) = 10/18 = 5/9

Second pick:

Number of white chocolate = 3

P(choosing a white chocolate) = 3 / 18 = 1/6

Hence, since both probabilities are independent :

P(choosing milk, then white chocolate) :

(5/9 × 1/6) = 5/54 = 0.0925 = 0.093

8 0

3 years ago

PLEASE HELP IMMEDIATELY!!

-Dominant- [34]

Answer:

Step-by-step explanation:

45-45-90

find hypotenuse: leg × root 2

$8 \times \sqrt{2}$

4 0

2 years ago