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3 years ago

Write an expression for: Five more sodas than twice the number of chips? And Daniela has 12 more books than Nicole?

Mathematics

1 answer:

Ahat [919]3 years ago

5 0

The expressions are.
5+2c
d=12+n

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Last summer, Marrero was at Newgrange, a stone mound in Ireland that has a diameter of 250 feet. He found a walkway 62 feet long

timofeeve [1]

Answer:

108 ft

Step-by-step explanation:

The line connecting the center with the end of the walkway is radius and is the hypotenuse of the right triangle.

Horizontal leg is the distance x.

Vertical leg has the length:

(250)/2 - 62 = 63 ft

Using Pythagorean theorem to find the length of x:

x² = (250/2)² - 63²
x² = 125² - 63²
x² = 11656
x = √11656
x = 107.96 ≈ 108

7 0

2 years ago

Do not understand please help Thank u

AysviL [449]

Answer:

43.4

Step-by-step explanation:

You will need to use the phythagoream therom, which is a² + b² = c². Plug the values in for the variables to get 7² + b² = 44², (c would equal the hypothenuse because the hypothenuse is the longest side of the triangle.) Square the values to get 49 + b² = 1936. Subtract 49 from both sides to get b² = 1887, then take the square root of both sides to get b = 43.4 (rounded to the nearest tenth.)

I am really sorry if I'm wrong, but have a good day/night! :)

6 0

2 years ago

Read 2 more answers

How to simplify the expressions of 5p-5-10-2p

Arada [10]

Answer:

3p - 15

Step-by-step explanation:

5p - 2p = 3p

-5 - 10 = -15

3p + - 15 = 3p - 15

4 0

2 years ago

PLZ ANSWER THIS thanksssss

just olya [345]

Answer:

It should be C) 11+ 3 5/8

Step-by-step explanation:

5 0

2 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