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2 years ago

Jessie saves 6 each week in how many weeks will she have saved at least 50

Mathematics

1 answer:

VLD [36.1K]2 years ago

7 0

Answer:

9 weeks

Step-by-step explanation:

Since you know your multiplication tables, you know that 6·8 = 48 and 6·9 = 54. This tells you 8 weeks is not enough, but Jessie will have saved at least 50 by week 9.

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Write 2.04 × 10 ⁴ as an ordinary number

ivann1987 [24]

Answer:

20400

Step-by-step explanation:

its 2.03x10x10x10x10 so first ten: 20.4 and ten: 204 3rd ten: 2040 last ten :20400

8 0

2 years ago

CONVERT TO SCIENTIFIC NOTATION:

Otrada [13]

1)1.082*10^9
2)1.496*10^8
3)2.2794*10^8
4)7.784*10^5
5)1.4236*10^9
6)2.867*10^9
7)4.4884*10^9
Hope it helped!

6 0

3 years ago

Ms. Larson was driving home from a vacation. At noon, she was 120 miles from home. Half an hour later, she was 98 miles from hom

melomori [17]

Answer: her speed was “44 miles per hour”

5 0

2 years ago

Use quadratic formula solve .7x^2 - 28x - 7

beks73 [17]

Answer:

x = 2 + sqrt(5) or x = 2 - sqrt(5)

Step-by-step explanation using the quadratic formula:

Solve for x over the real numbers:

7 (x^2 - 4 x - 1) = 0

Divide both sides by 7:

x^2 - 4 x - 1 = 0

Add 1 to both sides:

x^2 - 4 x = 1

Add 4 to both sides:

x^2 - 4 x + 4 = 5

Write the left hand side as a square:

(x - 2)^2 = 5

Take the square root of both sides:

x - 2 = sqrt(5) or x - 2 = -sqrt(5)

Add 2 to both sides:

x = 2 + sqrt(5) or x - 2 = -sqrt(5)

Add 2 to both sides:

Answer: x = 2 + sqrt(5) or x = 2 - sqrt(5)

6 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