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jeyben [28]
3 years ago
11

Problem

Mathematics
1 answer:
Contact [7]3 years ago
8 0

Answer:

<em>The width of the lid is 15 cm</em>

Step-by-step explanation:

<u>The Perimeter of a Rectangle</u>

Given a rectangle of width W and height H, the perimeter is calculated as:

P = 2(W + L)

Jon used 56 cm of ribbon to go around the edge of a rectangular lid. This means the perimeter is:

2(W + L) = 56

Dividing by 2:

W + L = 28

Since the lid was L=13 cm long, the width is:

W = 28 - 13 = 15

The width of the lid is 15 cm

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Answer:

True

Step-by-step explanation:

The reason I say this is because of the fact you can't go into retirement without knowing how much your going to get. You need to create a financial plan stating how much you will get and how much you have saved prior to this occurence.

Hope This Helps :)

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Find each probability below -- as a fraction, decimal and percent <br> Will give brainliest
Komok [63]

Answer:  

I was not sure to what decimal point to round it to

a) Fraction: 2/7    Decimal: 0.28571   Percent: 28.571 %

b) Fraction: 5/7    Decimal: 0.71428   Percent: 71.428 %

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What is 0.4% of 1000?
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0.4% = 0.04 
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Ostrovityanka [42]

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Step-by-step explanation:

5 0
3 years ago
i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.
Natali [406]

Answer:

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = 7

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = -5\sqrt 2

Step-by-step explanation:

Given

N = (-7,-1) --- terminal side of \varnothing

Required

Determine the values of trigonometric functions of \varnothing.

For \varnothing, the trigonometry ratios are:

sin\ \varnothing = \frac{y}{r}       cos\ \varnothing = \frac{x}{r}       tan\ \varnothing = \frac{y}{x}

cot\ \varnothing = \frac{x}{y}       sec\ \varnothing = \frac{r}{x}       csc\ \varnothing = \frac{r}{y}

Where:

r^2 = x^2 + y^2

r = \sqrt{x^2 + y^2

In N = (-7,-1)

x = -7 and y = -1

So:

r = \sqrt{(-7)^2 + (-1)^2

r = \sqrt{50

r = \sqrt{25 * 2

r = \sqrt{25} * \sqrt 2

r = 5 * \sqrt 2

r = 5 \sqrt 2

<u>Solving the trigonometry functions</u>

sin\ \varnothing = \frac{y}{r}

sin\ \varnothing = \frac{-1}{5\sqrt 2}

Rationalize:

sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

sin\ \varnothing = \frac{-\sqrt 2}{5*2}

sin\ \varnothing = \frac{-\sqrt 2}{10}

sin\ \varnothing = \frac{-1}{10}\sqrt 2

cos\ \varnothing = \frac{x}{r}

cos\ \varnothing = \frac{-7}{5\sqrt 2}

Rationalize

cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}

cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}

cos\ \varnothing = \frac{-7\sqrt 2}{10}

cos\ \varnothing = \frac{-7}{10}\sqrt 2

tan\ \varnothing = \frac{y}{x}

tan\ \varnothing = \frac{-1}{-7}

tan\ \varnothing = \frac{1}{7}

cot\ \varnothing = \frac{x}{y}

cot\ \varnothing = \frac{-7}{-1}

cot\ \varnothing = 7

sec\ \varnothing = \frac{r}{x}

sec\ \varnothing = \frac{5\sqrt 2}{-7}

sec\ \varnothing = \frac{-5}{7}\sqrt 2

csc\ \varnothing = \frac{r}{y}

csc\ \varnothing = \frac{5\sqrt 2}{-1}

csc\ \varnothing = -5\sqrt 2

3 0
3 years ago
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