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2 years ago

Help me to solve problem pless

Mathematics

2 answers:

sergejj [24]2 years ago

8 0

Answer:

5 x -3 +6

= 5 x 6 + -3

= 30 + -3

= 27

krok68 [10]2 years ago

7 0

Answer:

a) 15 c) -10 e) -18

Step-by-step explanation:

= 5 * (-3+6) = 5 * 3 = 15

= -10 * (7 - 6) = -10 * 1 = -10

= (-15 + 12) * 6 = -3*6 = -18

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Elsa saving money to buy a game so far she had saved $36 which is 3/4 of the total cost of the game how much does the game cost

natita [175]

Answer:

$48

Step-by-step explanation:

To figure out how much the game costs in total, we first need to divide $36 by 3 (3 is the numerator).

36 / 3 = 12

The number 12 is how much each quarter is worth as shown below.

1/4 = 12

2/4 = 24

3/4 = 36

4/4 = ?

To find the answer, all we do is add 12 to 36.

12 + 36 = $48

5 0

3 years ago

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Hi! I have a math question. This is a two part problem.

Mumz [18]

Answer:

Blair chosen to purchase shirt from company B. Given:

Each shirt costs 3.10$

Shipping cost is 4% (0.04) of total order.

=>Total cost without shipping: 150 x 3.10 = 465$

=> Total cost with 4% of shipping: 465 + 465 x 0.04 = 483.6$

=> Total cost Blair paid for one shirt: C = 483.6/150 = 3.22$

The markup is 75% of original cost:

=> Blair can multiply the total cost for one shirt by 0.75

Blair will sell shirt with price which is equal to the sum of total cost for one shirt and markup:

=> P = 3.22 + 3.22 x 0.75 = 5.64$

Hope this helps!

5 0

3 years ago

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Sever21 [200]

Answer:

same

Step-by-step explanation:

6 0

2 years ago

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What eigen value for this matix <br> (1 -2)<br> (-2 0)

natali 33 [55]

You find the eigenvalues of a matrix A by following these steps:

Compute the matrix $A' = A-\lambda I$ , where I is the identity matrix (1s on the diagonal, 0s elsewhere)
Compute the determinant of A'
Set the determinant of A' equal to zero and solve for lambda.

So, in this case, we have

$A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]$

The determinant of this matrix is

$\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4$

Finally, we have

$\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}$

So, the two eigenvalues are

$\lambda_1 = \dfrac{1+\sqrt{17}}{2},\quad \lambda_2 = \dfrac{1-\sqrt{17}}{2}$

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2 years ago

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Multiplying Powers with the Same Base

Luda [366]

Applying the product rule of exponents, each product of powers are matched with its simplified expression as:

1. $5 \times 5^3 = 5^4$

2. $5 \times 5^3 = 5^4$

3. $5^{-3} \times 5^{-3} = \frac{1}{5^6}$

4. $5^{-4} \times 5^{4} \times 5^0 = 5^0$

5. $5^{7} \times 5^{3} = 5^{10}$

To multiply the powers having the same base, we will apply the product rule for exponents.

<h3>What is the Product Rule for Exponents?</h3>

Base on the product rule for exponents, we have, $a^m \times a^n = a^{m + n} = a^{mn}$ .
In order to find the products of two given numbers that have the same base, the exponents would be added together.