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pochemuha
3 years ago
8

Simplify 5(h+8)^2+16=1486

Mathematics
1 answer:
natta225 [31]3 years ago
7 0
<span><span>h=7√<span><span>​6</span>​<span>​​</span></span></span>−8,−7<span>√<span><span>​6</span>​<span>​​</span></span></span>−8</span><span>​</span>
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I need the explanation and the answer of 14/40 in simplest form
Korvikt [17]

Answer:

\frac{7}{20}

Step-by-step explanation:

To simplify a fraction, you need to find the GCF (Greatest Common Factor) of the numerator (top number) and denominator (bottom number).

Factors of 14:

1, 2, 7, 14

Factors of 40:

1, 2, 4, 5, 8, 10, 20, 40

As you can see, 2 is the GCF of 14 and 40. So, we now divide both numbers by 2.

14 ÷ 2 = 7

40 ÷ 2 = 20

Therefore, \frac{14}{40} in simplest form is \frac{7}{20}

6 0
2 years ago
Numbers 7. 9 and 11.
Kay [80]

Answer:

7. Closed circle with the line going left (put dot on 8)

9. Closed circle on -9 with the line going right

11. Open circle on -4 1/2 with the line going left

Step-by-step explanation:

3 0
2 years ago
I need help with a math question that is 0.1x = 5/y
BabaBlast [244]
Unfortunately you have not included the directions.  Are you supposed to solve for x?  or for y?

Suppose the directions say, "Solve for x."  Then, clear out the decimal fraction by mult. both sides of the given equation by 10.  This will give you the solution, that is, a formula for x in terms of y.


3 0
3 years ago
It costs you $22.25 per day to rent a car in New York City. In addition, you have to pay 16 cents for every mile you travel. How
olganol [36]

Answer: $178.13

Step-by-step explanation:

418x0.16=66.88

22.25x5=111.25

111.25+66.88=178.13

4 0
2 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
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