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3 years ago

What is the simplest form of the square root of 1,764

Mathematics

2 answers:

garik1379 [7]3 years ago

6 0

The answer to this is 42.

olga55 [171]3 years ago

3 0

42 42X42=1764 <span>A square root is a number thet can be multiplied by two other numbers that are exactly the same like the square root of 16 is 4 because 4x4 is 16.</span>

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What is -4 3/10 - 8 3/10

nadya68 [22]

Answer:

-12 3/5

Step-by-step explanation:

-4 3/10 - 8 3/10

The signs are the same so we add and then take the sign

4 3/10 + 8 3/10

Add the whole numbers

4+8 = 12

Add the fractions

3/10 + 3/10 = 6/10

Simplify the fraction

6/10 = 3/5

Put them together

12 3/5

Add the sign

-12 3/5

4 0

2 years ago

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Write a question that represents the following equation. 5 · n = 35. plssss help hurry

snow_lady [41]

Answer:

what is the value of {n}to get product 35?

Step-by-step explanation:

(n is seven by the way!!)

6 0

2 years ago

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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

You deposit $4300 in an account that earns 2% per year simple interest. The equation that represents this situation is: A(n) = 4

Likurg_2 [28]

Answer: $4902

Step-by-step explanation:

Amount = P (1 + rt)

= 4300 (1 + .02 * 7)

= 4300 (1 + .14)

= 4300 (1.14)

= 4902

3 0

3 years ago

X^2-4x+1=0 . RESOLVE PLS

Mkey [24]

$x^2-4x+1=0\\ x^2-4x+4-3=0\\ (x-2)^2=3\\ x-2=\sqrt 3\vee x-2=-\sqrt3\\ x=2+\sqrt 3\vee x=2-\sqrt3$

4 0

3 years ago