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2 years ago

7. 56 + 10 = 75 - Any one know how to solve this problem

Mathematics

1 answer:

bonufazy [111]2 years ago

4 0

Answer:

66=75

(false)

Step-by-step explanation:

56+10=75

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Whoever answer is the best gets brainlist

Otrada [13]

The answer is 18^8 because u add exponents

8 0

3 years ago

Read 2 more answers

Suppose a bank charges a base rate of $5 plus $0.10 per check for your checking account. You can switch to a different account t

wolverine [178]

Answer:

10 number of checks.

Cost = $6

Step-by-step explanation:

Given that:

Fixed Charges of first bank = $5

Charges of first bank per check = $0.10

Fixed Charges of second bank = $4

Charges of second bank per check = $0.20

To find:

Number of checks such that the charges for the banks become the same?

Solution:

Let the number of checks = $x$

Cost for the first bank for $x$ checks = $5 + $0.10 $x$

Cost for the second bank for $x$ checks = $4 + $0.20 $x$

As per question statement, both the costs are the same.

Comparing the values:

$5 + 0.10x = 4 + 0.20x\\\Rightarrow 0.10x = 1\\\Rightarrow x = 10$

So, for 10 number of checks the cost will be same.

The cost = 4 + 0.20 $\times$ 10 = <em>$6</em>

8 0

3 years ago

Jack said that if you take a number and multiply it by a fraction, the product will always be smaller than what you started with

ELEN [110]

Answer:

Is not correct. The product not always will be smaller.

Step-by-step explanation:

A fraction could increase a number if you multiply it.

If the numerator is higher to the denominator, it will increase it.

If the numerator is smaller to the denominator, it will decrease it.

<u>For example,</u>

10 x 4/3

For fraction multiplication, multiply the numerators and then multiply the denominators to get

10×4/1×3=40/3= 13,33

13,33 is higher than 10.

5 x 10/3

For fraction multiplication, multiply the numerators and then multiply the denominators to get

5×10/1×3=50/3

16,66 is higher than 5.

5 0

3 years ago

Let y 00 + by0 + 2y = 0 be the equation of a damped vibrating spring with mass m = 1, damping coefficient b > 0, and spring c

stira [4]

Answer:

Step-by-step explanation:

Given that:

The equation of the damped vibrating spring is y" + by' +2y = 0

(a) To convert this 2nd order equation to a system of two first-order equations;

let y₁ = y

y'₁ = y' = y₂

So;

y'₂ = y"₁ = -2y₁ -by₂

Thus; the system of the two first-order equation is:

y₁' = y₂

y₂' = -2y₁ - by₂

(b)

The eigenvalue of the system in terms of b is:

$\left|\begin{array}{cc}- \lambda &1&-2\ & -b- \lambda \end{array}\right|=0$

$-\lambda(-b - \lambda) + 2 = 0 \ \\ \\\lambda^2 +\lambda b + 2 = 0$

$\lambda = \dfrac{-b \pm \sqrt{b^2 - 8}}{2}$

$\lambda_1 = \dfrac{-b + \sqrt{b^2 -8}}{2} ; \ \lambda _2 = \dfrac{-b - \sqrt{b^2 -8}}{2}$

(c)

Suppose $b > 2\sqrt{2}$ , then λ₂ < 0 and λ₁ < 0. Thus, the node is stable at equilibrium.

(d)

From λ² + λb + 2 = 0

If b = 3; we get

$\lambda^2 + 3\lambda + 2 = 0 \\ \\ (\lambda + 1) ( \lambda + 2 ) = 0\\ \\ \lambda = -1 \ or \ \lambda = -2 \\ \\$

Now, the eigenvector relating to λ = -1 be:

$v = \left[\begin{array}{ccc}+1&1\\-2&-2\\\end{array}\right] \left[\begin{array}{c}v_1\\v_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]$

$\sim v = \left[\begin{array}{ccc}1&1\\0&0\\\end{array}\right] \left[\begin{array}{c}v_1\\v_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]$

Let v₂ = 1, v₁ = -1

$v = \left[\begin{array}{c}-1\\1\\\end{array}\right]$

Let Eigenvector relating to λ = -2 be:

$m = \left[\begin{array}{ccc}2&1\\-2&-1\\\end{array}\right] \left[\begin{array}{c}m_1\\m_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]$

$\sim v = \left[\begin{array}{ccc}2&1\\0&0\\\end{array}\right] \left[\begin{array}{c}m_1\\m_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]$

Let m₂ = 1, m₁ = -1/2

$m = \left[\begin{array}{c}-1/2 \\1\\\end{array}\right]$

∴

$\left[\begin{array}{c}y_1\\y_2\\\end{array}\right]= C_1 e^{-t} \left[\begin{array}{c}-1\\1\\\end{array}\right] + C_2e^{-2t} \left[\begin{array}{c}-1/2\\1\\\end{array}\right]$

So as t → ∞

$\mathbf{ \left[\begin{array}{c}y_1\\y_2\\\end{array}\right]= \left[\begin{array}{c}0\\0\\\end{array}\right] \ \ so \ stable \ at \ node \ \infty }$

5 0

2 years ago

The width of a rectangle is 5 cm less than the length. The perimeter is 38 cm. Find the dimensions of the rectangle

VikaD [51]

Answer:

Step-by-step explanation:

38=2(5)+2(x ) find X

2x5-10

38-10=28

28/2=14

14/2=7

x =7

so you length is 7

8 0

3 years ago