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3 years ago

9

A rectangular garden has a total area of 48 yards.Draw and label two possible rectangular gardens with diffrent side lenghts tha

t have the same area.

1 answer:

Vikki [24]3 years ago

7 0

Answer:

possible answers are

6x8

4x 12

2x 24

Step-by-step explanation:

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The temperature of a cooling liquid over time can be modeled by the exponential function T(x) = 60(1/2)^(x/30) + 20, where T(x)

Alexus [3.1K]

Based on the scenario, the initial temperature would be :

28 = (60) (1/2)^x/30 + 20

(60) (1/2)^0/30 + 20 = 60 + 20 =

80 C

Hope this helps

6 0

3 years ago

Give four rational numbers equivalent to : a) 2/7 b) 3/5

serg [7]

Answer:

gg

Step-by-step explanation:

6 0

2 years ago

3.<br> QR<br> find the arc length<br> 02.83<br> 021.99<br> O 12.57<br> 0 34.56

jeka94

Your answer should be 12.57

8 0

2 years ago

2. Which equation is equivalent to 2 + 4x = 5 - 3(x + 1)? A. 2 + 4x = 2x - 3 B. 2 + 4% = 5 - 3 -3. C. 6x = 5-(+ 1). D. 2 + 4 = 2

Andre45 [30]

Answer:

B) 2 + 4x = 5 - 3 - 3x

Step-by-step explanation:

By simplifying JUST the right side and re-arranging them

2 + 4x = 5 - 3(x + 1),

2 + 4x = 5 - 3x -3,

2 + 4x = 5 -3 - 3x

3 0

3 years ago

Let <img src="https://tex.z-dn.net/?f=i" id="TexFormula1" title="i" alt="i" align="absmiddle" class="latex-formula"> be the imag

VLD [36.1K]

$Hey~freckledspots!\\----------------------$

$We~will~solve~for~i^{425}!$

$Rule~of~exponent: a^{b + c} = a^ba^c\\Apply:~i^{425}~=~i^{424}i\\ \\Rule~of~exponent: a^{bc} = (a^{b})^c\\Apply: i^{424} = i(i^2)^{212} \\\\Rule~of~imaginary~number: i^2 = -1\\Apply: i(i^2)^{212} = -1^{212}i\\\\Rule~of~exponent~if~n~is~even: -a^n = a^n\\Apply: -1^{212}i = 1^{212}i\\\\Simplify: 1^{212}i = 1i\\Multiply: 1i * 1 = i\\----------------------\\$

$Now~let's~solve~1^{14}!\\\\Rule~of~exponent: a^{b + c} = a^ba^c\\Apply: i^{14} = (i^2)^7\\\\Rule~of~imaginary~number: i^2 = -1\\Apply: (i^2)^7 = -1^7\\\\Rule~of~exponent~if~n~is~odd: (-a)^n = -a^n\\Apply: -1^7 = -1^7\\\\Simplify: -1^7 = -1\\----------------------\\Now,~we~have: i-1+i^{-14}+i^{44}\\----------------------$

$Now~lets~solve~i^{-14}\\\\Rule~of~exponent: a^{-b} = \frac{1}{a^b} \\Apply: i^{-14} = \frac{1}{i^{14}} \\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: \frac{1}{i^{14}} = \frac{1}{(i^2)^7}\\ \\Rule~of~imagianry~number: i^2 = -1\\Apply: \frac{1}{(i^2)^7} = \frac{1}{-1^7} \\\\Simplify: \frac{1}{-1^7} = \frac{1}{-1} \\\\Rule~of~fractions: \frac{a}{-b} = -\frac{a}{b} \\Apply: \frac{1}{-1} = -\frac{1}{1} = -1\\----------------------\\Now,~we~have: i-1-1+i^44\\----------------------$

$Now~let's~solve~i^{44}!\\\\Rule~of~exponent: a^{bc} = (a^b)^c\\Apply: i^{44} = (i^2)^{22}\\\\Rule~of~imaginary~numbers: i^2 = -1\\Apply: (i^2)^{22} = -1^{22}\\\\Rule~of~exponent~if~n~is~even: (-a)^n = a^n\\Apply: -1^{22} = 1^{22}\\\\Simplify: 1^{22} = 1\\----------------------\\Now,~we~have~i-1-1+1\\----------------------$

$Now~let's~simplify~the~expression!\\\\= i-1-1+1 \\= 1 + i -2\\= -1+i\\----------------------$

$Answer:\\\large\boxed{-1+i}\\----------------------$

$Hope~This~Helped!~Good~Luck!$

8 0

3 years ago

Read 2 more answers