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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X^2 + __x + 49 pls i need help and i need 4x^2-24x+__

nadezda [96]

Answer:

4x2−24x4x2-24x

Factor 4x4x out of 4x24x2.

4x(x)−24x4x(x)-24x

Factor 4x4x out of −24x-24x.

4x(x)+4x(−6)4x(x)+4x(-6)

Factor 4x4x out of 4x(x)+4x(−6)4x(x)+4x(-6).

4x(x−6)4x(x-6)

4x2−24x4x2-24x

Step-by-step explanation:

tried

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oee [108]

Yes they are equal they both have a side that is 25 degrees

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3 years ago

Find the value of the constant m √150 - √12m + √54 = 0

Tomtit [17]

Answer:

m is √2

Step-by-step explanation:

${ \tt{ \sqrt{150} - \sqrt{12}m + \sqrt{54} = 0 }} \\ { \tt{ \sqrt{12}m } = \sqrt{150} - \sqrt{54} } \\ { \tt{ (\sqrt{4 \times 3}) m = ( \sqrt{25 \times 6} ) - ( \sqrt{9 \times 6}) }} \\ { \tt{( \sqrt{4 \times 3} )m = 5 \sqrt{6} - 3 \sqrt{6} }} \\ { \tt{2 \sqrt{3}m = 5 \sqrt{6} - 3 \sqrt{6} }} \\ { \tt{2 \sqrt{3} m = 2 \sqrt{6} }} \\ { \tt{ \sqrt{3} m = \sqrt{6} }} \\ { \tt{ \sqrt{3} m = ( \sqrt{3} \times \sqrt{2} ) }} \\ { \tt{m = \sqrt{2} }}$