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3 years ago

How do you find y? ∣3y−1∣+∣2y+3∣>5

1 answer:

5 0

Let's solve your inequality step-by-step.

3y−1+2y+3>5

Step 1: Simplify both sides of the inequality.5y+2>5

Step 2: Subtract 2 from both sides.
5y+2−2>5−2
5y>3
Step 3: Divide both sides by 5.
5y/5>3/5
y>3/5

hopefully this helps and here's a link that i used to use a lot in middle school
https://www.symbolab.com/solver/solve-for-equation-calculator

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How tall does an adult male need to be in order to be assigned to the “tall” category?

Naddik [55]

53.07 inches is the answer i found

6 0

3 years ago

Lawrence sells 40% of the apples on his cart, and are left with 420 apples. How many apples had Lawrence started with?

lutik1710 [3]

Answer:

700 apples

Step-by-step explanation:

Let

x = Total number of apples in the cart

Percentage of apples sold = 40%

Number of apples sold = Percentage of apples sold * Total number of apples in the cart

= 40% * x

= 0.4 * x

= 0.4x

Number of apples left in the cart = 420

Total number of apples in the cart = Number of apples sold + Number of apples left in the cart

x = 0.4x + 420

x - 0.4x = 420

0.6x = 420

x = 420/0.6

x = 700

x = Total number of apples in the cart = 700

Lawrence started with 700 apples

4 0

3 years ago

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$