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3 years ago

Help me u get brainliest plus more points :D

Mathematics

1 answer:

vredina [299]3 years ago

7 0

Answer:

6 is 56.25

7 is 97.50

Step-by-step explanation:

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Which of the following is a possible situation for the graph shown?

docker41 [41]

Step-by-step explanation:

but the graph is not given here

4 0

2 years ago

6/8 divide by 8/12 A. 9/2 B. 8/9 C. 9/8 D. 9/32

Musya8 [376]

Answer:

C. 9/8

Step-by-step explanation:

$\frac{6}{8} \div \frac{8}{12}$

Use KCF which means Keep the first fraction, Change the second fraction by flipping it and change the division sign to a multiplication symbol:

$\frac{6}{8} \times \frac{12}{8}$

multiply the numerator together:

6 × 12 = 72

multiply the denominator together:

8 × 8 = 64

now the fraction is:

$\frac{72}{64}$

simplify further by dividing both numbers by 8:

72 ÷ 8 = 9

64 ÷ 8 = 8

so the fraction is 9/8

3 0

3 years ago

a college employs 15 stewards. what is the average monthly wage of the stewards if the college pays 226.20 naira in wages each m

Kitty [74]

What are the options

6 0

3 years ago

Can someone help me me answer this question please and thankyou!??

Oksi-84 [34.3K]

12 + 2 - (4 × 5^2) ÷ 7 + 1

12 + 2 - (4 × 25) ÷ 7 + 1

12 + 2 - 100 ÷ 7 + 1

12+2-100/7+1

15-100/7

Equal 5/7
Rise over run 5 is rise and 7 is run

5 0

3 years ago

Write the linear system of differential equations in matrix form then solve the system.

zvonat [6]

In matrix form, the system is

$\dfrac{\mathrm d}{\mathrm dt}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1&1\\4&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$

First find the eigenvalues of the coefficient matrix (call it $\mathbf A$ ).

$\det(\mathbf A-\lambda\mathbf I)=\begin{vmatrix}1-\lambda&1\\4&1-\lambda\end{vmatrix}=(1-\lambda)^2-4=0\implies\lambda^2-2\lambda-3=0$

$\implies\lambda_1=-1,\lambda_=3$

Find the corresponding eigenvector for each eigenvalue:

$\lambda_1=-1\implies(\mathbf A+\mathbf I)\vec\eta_1=\vec0\implies\begin{bmatrix}2&1\\4&2\end{bmatrix}\begin{bmatrix}\eta_{1,1}\\\eta_{1,2}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$

$\lambda_2=3\implies(\mathbf A-3\mathbf I)\vec\eta_2=\vec0\implies\begin{bmatrix}-2&1\\4&-2\end{bmatrix}\begin{bmatrix}\eta_{2,1}\\\eta_{2,2}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$