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2 years ago

A rectangle has a length of 12 and a width of 9. What is the

Mathematics

1 answer:

choli [55]2 years ago

3 0

Answer:

Step-by-step explanation:

9 squared + 12 squared

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Write x2 - 8x + 13 = 0 in the form (x - a)2 = b, where a and b are integers. a.)(x - 4)2 = 3 b.) (x - 3)2 = 2 c.)(x - 2)2 = 1 d.

loris [4]

The answer is a.)(x - 4)2 = 3.

(x - a)² = b can be expressed as:
x² - 2ax + a² = b ⇒ x² - 2ax + a² - b = 0

Our equation is x² - 8x + 13 = 0.
The general formula is x² - 2ax + (a² - b) = 0

Thus:
8x = 2ax and a² - b = 13

Divide both sides of the first equation (8x = 2ax) by x:
8 = 2a ⇒ a = 8 ÷ 2 = 4

Replace a in the second equation (a² - b = 13):
4² - b = 13 ⇒ b = 4² - 13 = 16 - 13 = 3

Now when we have a and b, let's just replace them in the general equation:
(x - a)² = b
(x - 4)² = 3

4 0

3 years ago

Bagels: six in a bag Apples: eight in a bag Cookies: twelve in a box Juice Boxes: nine in a box Find the least number of package

morpeh [17]

Bagels 6x12=72
apples 8x9=72
cookies 12x6=72
juice 9x8=72

72/4 kids is 18 lunches

7 0

3 years ago

Which number is irrational 25/9

pantera1 [17]

We have a fraction of two integers, so 25/9 is rational. A rational number is any fraction of two integers (as long as the denominator is not zero).

5 0

3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height

Crank

Answer:

a) h = 0.1: $\bar v = -11\,\frac{ft}{s}$ , h = 0.01: $\bar v = -10.1\,\frac{ft}{s}$ , h = 0.001: $\bar v = -10\,\frac{ft}{s}$ , b) The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

Step-by-step explanation:

a) We know that $y = 30\cdot t -10\cdot t^{2}$ describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity ( $\bar v$ ), measured in feet per second, can be done by means of the following definition:

$\bar v = \frac{y(2+h)-y(2)}{h}$

Where:

$y(2)$ - Position of the ball evaluated at $t = 2\,s$ , measured in feet.

$y(2+h)$ - Position of the ball evaluated at $t =(2+h)\,s$ , measured in feet.

$h$ - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

$h = 0.1\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}$

$y(2.1) = 18.9\,ft$

$\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}$

$\bar v = -11\,\frac{ft}{s}$

$h = 0.01\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}$

$y(2.01) = 19.899\,ft$

$\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}$

$\bar v = -10.1\,\frac{ft}{s}$

$h = 0.001\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}$

$y(2.001) = 19.99\,ft$

$\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}$

$\bar v = -10\,\frac{ft}{s}$

b) The instantaneous velocity when $t = 2\,s$ can be obtained by using the following limit:

$v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}$

$v(t) = \lim_{h \to 0} 30-20\cdot t-10\cdot h$

$v(t) = 30\cdot \lim_{h \to 0} 1 - 20\cdot t \cdot \lim_{h \to 0} 1 - 10\cdot \lim_{h \to 0} h$

$v(t) = 30-20\cdot t$

And we finally evaluate the instantaneous velocity at $t = 2\,s$ :

$v(2) = 30-20\cdot (2)$

$v(2) = -10\,\frac{ft}{s}$

The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

8 0

3 years ago

Particle 1 of charge q1 �� 5.00q and particle 2 of charge q2 �� 2.00q are fixed to an x axis. (a) as a multiple of distance

Naya [18.7K]

Assuming that the particle is the 3rd particle, we know that it’s location must be beyond q2; it cannot be between q1 and q2 since both fields point the similar way in the between region (due to attraction). Choosing an arbitrary value of 1 for L, we get

k q1 / d^2 = - k q2 / (d-1)^2

Rearranging to calculate for d:

(d-1)^2/d^2 = -q2/q1 = 0.4 
 d^2-2d+1 = 0.4d^2 
0.6d^2-2d+1 = 0
d = 2.72075922005613
d = 0.612574113277207

We pick the value that is > q2 hence,

d = 2.72075922005613*L

d = 2.72*L

3 0

3 years ago