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3 years ago

The net force acting an object A is The net force acting an object B is

Mathematics

2 answers:

Lynna [10]3 years ago

8 0

Answer:

A: 6N to the right

B: 7N to the right

Step-by-step explanation:

im not sure what the first option is for both questions and im not sure if this will answer ur question but i hope this helps

krok68 [10]3 years ago

4 0

Answer:

Step-by-step explanation:

yes it does it take for the time being a new one if u are not going to be proud of u for doing this for my younger sister can't be bothered doing this for my younger sister can't be bothered doing this for my younger sister can't be bothered doing it is negative for a cat that does he cover per with an open interest of these are right or wrong lol but I'm cool with this asapp and the

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Does anyone know how to turn these fractions into percents??? please help!!!

belka [17]

You divide the fractions to get percents

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3 years ago

The base of a rectangular prism is 5.4 centimeters long and 4.9 centimeters wide. It is 3.8 centimeters tall. What is the volume

lorasvet [3.4K]

So, We Need To Examine The Problem. So, We Know That We Need To Find The Volume Of A Rectangular Prism. We Also Know That The Dimensions Are 4.9 • 3.8 • 5.4.
So, We Need To Remember The Formula For Volume Of A Rectangular Prism.
V = B • W • H
So, we need to plug in the known values.
V = 4.9 • 3.8 • 5.4.
So, Lets Solve.
4.9 • 3.8 = 18.62
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3 years ago

Which expression is equivalent to 3x+10−x+12 a.24x b.4x + 22 c.26x d.2x + 22

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8 0

4 years ago

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 $-4a^3b$ .

3 0

3 years ago

What's 49 divided by 261

Tpy6a [65]

.19
Hope this helped!

8 0

3 years ago

Read 2 more answers