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2 years ago

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The conversion rates between kilograms and pounds are: 1 k g ≈ 2.2 l b s 1 l b ≈ 0.454 k g A. How many kilograms is a backpack t

hat weighs 6 pounds? 2.724 B. How many pounds is a cat that weighs 11 kilograms?

1 answer:

denis-greek [22]2 years ago

3 0

Answer:

$6\ lb = 2.724\ kg$

$11\ kg = 24.2\ lb$

Step-by-step explanation:

Given

$1\ kg = 2.2\ lb$

$1\ lb = 0.454\ kg$

Required

- Convert 6 lb to kg

- Convert 11kg to lb

To solve for (a), we make use of the second conversion unit:

$1\ lb = 0.454\ kg$

Multiply both sides by 6

$6 * 1\ lb = 0.454\ kg * 6$

$6\ lb = 2.724\ kg$

<em>Hence, a 6lb backpack weighs 2.724kg</em>

Solving (b): 11 kg to lb

Here, we make use of the first conversion unit

$1\ kg = 2.2\ lb$

Multiply both sides by 11

$11 * 1\ kg = 2.2\ lb * 11$

$11\ kg = 24.2\ lb$

<em>Hence, the cat weighs 24.2lb</em>

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Find the length of the indicated side in each triangle, to nearest unit.

gogolik [260]

Answer:

A= 56

E= 59

Step-by-step explanation:

All triangle are 180 degrees so

A.

71 + 53 = 124

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2 years ago

Michelle plans on raising $80 for her softball team. To raise the money, she is selling team shirts for $10 each.

8_murik_8 [283]

Given:

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So far she has raised $20.

To find:

The number of more shirts she needs to sell.

Solution:

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Michelle plans on raising $80. So,

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Subtract both sides by 20.

$10x=80-20$

$10x=60$

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2 years ago

Please help me answer this question

ratelena [41]

By the definition of the hyperbolic function tanh x, we have proven that $\frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}$

<h3>Hyperbolic functions & Proof of identities </h3>

By definition

$tanh \ x=\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}}$

Then,

$\frac{1-tanh \ x}{1 + tanh \ x}=\frac{1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }{1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }$

$=1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} \div (1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} })$

$=\frac{e^{x} +e^{-x}-(e^{x} -e^{-x})}{e^{x} +e^{-x}} \div (\frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} })$

$=\frac{e^{x} +e^{-x}-e^{x} +e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} }$

$=\frac{e^{-x}+e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{x} }{e^{x} +e^{-x}} }$

$=\frac{2e^{-x}}{e^{x} +e^{-x}} \div \frac{2e^{x} }{e^{x} +e^{-x}} }$

$=\frac{2e^{-x}}{e^{x} +e^{-x}} \times \frac{e^{x} +e^{-x}}{2e^{x}}$

$=\frac{2e^{-x}}{1} \times \frac{1}{2e^{x}}$

$=\frac{2e^{-x}}{2e^{x}}$

$=\frac{e^{-x}}{e^{x}}$
$=e^{-x} \times \frac{1}{e^{x}}$

$= e^{-x} \times e^{-x}$

$= e^{-x+-x}$

$= e^{-x-x}$

$= e^{-2x}$

Hence, we have proven that $\frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}$

Learn more on Proof of Identities here: brainly.com/question/2561079

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2 years ago

How do you solve a question algbraically

Julli [10]

Answer:

Step 1: Multiply the entire first equation by 2.

Step 2: Rewrite the system of equations, replacing the first equation with the new equation.

Step 3: Add the equations.

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3 years ago

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Nikitich [7]

Answer:

Step-by-step explanation:

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3 years ago