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3 years ago

Please help! combine like terms to create an equip expression 0.4m - 0.8 - 0.8m

Mathematics

1 answer:

soldi70 [24.7K]3 years ago

4 0

Answer:

1.2-0.8

Step-by-step explanation:

Add 0.4 and 0.8

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4st^2-s^2 were s=5 and t=7

barxatty [35]

➣ Assignment: $\bold{Solve \ Equation: \ \left(4\left(5\cdot \:7^2\right)\right)-5^2}$

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➣ Answer: $\boxed{\bold{955}}$

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Explanation: ↓↓↓↓↓

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➤ [ Step One ] Follow PEMDAS Order Of Operations

$\bold{4\left(5\cdot \:7^2\right): \ 980}$

➤ [ Step Two ] Rewrite Equation

$\bold{4\left(5\cdot \:7^2\right)}$

➤ [ Step Three ] Calculate Within Parenthesis

$\bold{4\left(5\cdot \:7^2\right): \ 245}$

➤ [ Step Four ] Rewrite Equation

$\bold{4\cdot \:245}$

➤ [ Step Five ] Multiply

$\bold{4\cdot \:245: \ 980}$

➤ [ Step Six ] Rewrite Equation

$\bold{980-5^2}$

➤ [ Step Seven ] Calculate Exponents

$\bold{5^2: \ 25}$

➤ [ Step Eight ] Rewrite Equation

$\bold{980-25}$

➤ [ Step Nine ] Solve

$\bold{955}$

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$\bold{\rightarrow Mordancy \leftarrow}$

4 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]

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

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

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

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

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

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

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

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

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

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What Value Does Point P Represent On The Number Line Shown Here?

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The value is -3.2. Hope it helps!

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3 years ago

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A certain college graduate borrows 5510 dollars to buy a car. The lender charges interest at an annual rate of 17%. Assuming tha

Diano4ka-milaya [45]

Answer:

His annual payment rate required to pay off the loan in 7 years is $k = $2362.41$ a year.

He has to pay $11026.85 in interest

Step-by-step explanation:

The first step to solve this problem is find how much the student will have to pay for the car. So, we have to find how much the present value of the car(in $) will be worth in 7 years. This is a compound interest problem:

The compound interest formula is given by:

$A = P(1 + \frac{r}{n})^{nt}$

Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.

In this exercise, we have:

A = The value we want to find

P = The initial value = 5510

r = 0.17

n = 1

t = 7

$A = 5510(1 + \frac{0.17}{1})^{7}$