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3 years ago

Write a fraction that is equivalent to 3/5 that has a denominator of 20.

Mathematics

1 answer:

jek_recluse [69]3 years ago

6 0

Answer:

12/20

Step-by-step explanation:

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Please simplify with a explanation!~

PilotLPTM [1.2K]

Answer:

-4y + 20

Step-by-step explanation:

4(-y+5)

What we need to do here is get rid of the parentheses, which can be achieved through multiplication.

This is another way of saying 4 times -y plus 4 times 5.

Let's break the the parentheses down into 2 separate formulas.

4(-y) and 4(5)

Now we can multiply these two together to get -4y and +20. When combined, we get -4y +20.

6 0

3 years ago

Can anyone help me with this

Likurg_2 [28]

7/13 = about 0.54
-1 1/2 = -1.5

The answer is
-1.7, -1 1/2, 0.04, 0.1, 7/13

5 0

3 years ago

Read 2 more answers

HELP! I will mark branliest, please answer this question!!!<br><br> The value of x is ____?

aniked [119]

45x+angle trs = 180 straight line

angle trs = 180-45x

angle trs + angle t + angle tsr = 180 (sum of angles in a triangle =180)

(180-45x) + 25x + 57+x = 180

combine like terms

180 +57 -45x+25x+x=180

237-19x=180

subtract 237 from each side

-19x =-57

divide by -19

x=3

4 0

3 years ago

Can someone please help me, please

e-lub [12.9K]

I can only help you if you give me a lot of points

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