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2 years ago

10

A basketball is 11 times as heavy as a tennis ball. If a tennis ball weighs 58 grams, what is the weight of a basketball in kilo

grams

1 answer:

Vlada [557]2 years ago

8 0

Answer:

0.638 KGs or 638 grams

Step-by-step explanation:

as we know 1000grams = 1 kilogram

therefore the weight of basketball :

11*58/1000 = 0.638 KGs

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(a + b)(a ² - ab + b ²)

erastova [34]

Hi there!

You can use distributive property to solve this equation.

These are the list of terms you have to multiply :
a × a² = a³
a × -ab = -a²b
a × b² = ab²
b × a² = a²b
b × -ab = -ab²
b × b² = b³

add all of those terms...
a³ - a²b + ab² +a²b -ab² + b³
simplify by adding like terms... (- a²b + a²b , ab² - ab²)

a³ + b³

Hope this helped!! :)

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2 years ago

Please answer this and I’ll make ya

Fynjy0 [20]

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3 years ago

Just how much is a million?

Fudgin [204]

Hey You!

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2 years ago

Read 2 more answers

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

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2 years ago

Helps pls :-/<br> number 12<br> any help is greatly appreciated!

Alexandra [31]

X = 5, that’s the answer to question 12

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3 years ago