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nydimaria [60]
2 years ago
14

What are the type of Forces​

Mathematics
1 answer:
Brut [27]2 years ago
6 0

pull and push are the two basic kinds of forces then there is gravity and downrange and friction

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Combine the following fractions by addition or subtraction as directed.
Marina86 [1]
Answer:\frac{6}{y^2-xy}-\frac{6}{x^2-xy}=\frac{6(x+y)}{xy(y-x)}

Explanation:

Combining the fractions, we have:

\begin{gathered} \frac{6(x^2-xy)-6(y^2-xy)}{(x^2-xy)(y^2-xy)} \\  \\ =\frac{6x^2-6xy-6y^2+6xy}{(x^2-xy)(y^2-xy)} \\  \\ =\frac{6x^2-6y^2}{x(x-y).y(y-x)} \\  \\ =\frac{6(x-y)(x+y)}{-xy(x-y)^2} \\  \\ =\frac{-6(x+y)}{xy(x-y)} \\  \\ =\frac{6(x+y)}{xy(y-x)} \end{gathered}

8 0
1 year ago
Multiply.4 3/10× 2 1/3Enter your answer in the box as a mixed number in simplest form.
faust18 [17]

Answer:

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Step-by-step explanation:

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4 0
2 years ago
Charity invests $1891 into a savings account. The account will accrue 9% interest to be compounded semiannually. Provided she ma
Norma-Jean [14]

Answer:

B) 12,010.9

Step-by-step explanation:

Set up an equation:

A=1891(1+\frac{0.09}{2})^{2*21}

Solve:

A=1891(1.045)^{42}

A=12010.90488

8 0
2 years ago
If f(x) = x over 2 + 8, what is f(x) when x = 10? 4 9 13 36
Katena32 [7]
If x=10 than f(x)=1 because 2+8=10/10 =1

5 0
2 years ago
Read 2 more answers
If the two lines below are perpendicular and the slope ofthe red line is -, what is the slope of the green line?51015-5-5-105
inysia [295]

If two lines are perpendicular, then the product of their slopes is -1:

m_r\cdot m_g=-1...(1)

From the problem, we identify:

m_r=-\frac{4}{3}\text{  \lparen the slope of the red line\rparen}

Then, using the equation (1), we calculate the slope of the green line:

\begin{gathered} -\frac{4}{3}\cdot m_g=-1 \\  \\ \therefore m_g=\frac{3}{4} \end{gathered}

Answer: Option D

4 0
1 year ago
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