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nataly862011 [7]
3 years ago
7

SOLVE. (5n)÷(30m)+(2m+4n)÷(30m)

Mathematics
2 answers:
Andru [333]3 years ago
6 0
Note, since both factors have the same denominator, we can add the numerators together.

Numerator :

5n + (2m + 4n) = 5n + 2m + 4n = 9n + 2m

Denominator :

30m

Thus,

= (9n + 2m) / 30m
romanna [79]3 years ago
6 0
(9n+2m)/30m

~Hope This Helps :)
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Need help with this math question 30 points correct answers only pls
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its the third answer

Step-by-step explanation:

and its only 5 points not 30

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The number 3456 is divisible by wich single-digit number?
Gnesinka [82]

I'm assuming what you mean is what can you divide 3456 by to get a natural number.

First of all every number is divisible by 1.  

Every even number is divisible by 2, so since 3456 is an even number, it can be divided by 2.  That would equal 1728.

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You can also divide it by 4, to get 865.

Divided by 6, it's 576.

Divided by 8, it's 432.

And lastly, 3456 divided by 9 is 384.

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Hope that helps!

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A basketball court is in the shape of a rectangle. It has a length of (x − 7) meters and a width of (x + 3) meters. The expressi
Step2247 [10]
X^2 -3x+7x-21

so simplified is:

x^2 - 4x - 21


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Which graph best represents y=−2x+3
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Step-by-step explanation:

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Amy is pulling a wagon with a force of 30 pounds up a hill at an angle of 25°. Give the force exerted on the wagon as a vector a
Fofino [41]

Answer:

Vector (ordered pair - rectangular form)

\vec F = (27.189,12.679)\,[lbf]

Vector (ordered pair - polar form)

\vec F = (30\,lbf, 25^{\circ})

Sum of vectorial components (linear combination)

\vec F = 27.189\cdot \hat{i} + 12.679\cdot \hat{j}\,[N]

Step-by-step explanation:

From statement we know that force exerted on the wagon has a magnitude of 30 pounds-force and an angle of 25° above the horizontal, which corresponds to the +x semiaxis, whereas the vertical is represented by the +y semiaxis.

The force (\vec F), in pounds-force, can be modelled in two forms:

Vector (ordered pair - rectangular form)

\vec F =  \left(\|\vec F\|\cdot \cos \theta, \|\vec F\|\cdot \sin \theta\right) (1)

Vector (ordered pair - polar form)

\vec F = \left(\|\vec F\|, \theta\right)

Sum of vectorial components (linear combination)

\vec {F} = \left(\|\vec F\|\cdot \cos \theta\right)\cdot \hat{i} + \left(\|\vec F\|\cdot \sin \theta \right)\cdot \hat{j} (2)

Where:

\|\vec F\| - Norm of the vector force, in newtons.

\theta - Direction of the vector force with regard to the horizontal, in sexagesimal degrees.

\hat{i}, \hat{j} - Orthogonal axes, no unit.

If we know that \|\vec F\| = 30\,lbf and \theta = 25^{\circ}, then the force exerted on the wagon is:

Vector (ordered pair - rectangular form)

\vec F= \left(30\cdot \cos 25^{\circ}, 30\cdot \sin 25^{\circ}\right)\,[lbf]

\vec F = (27.189,12.679)\,[lbf]

Vector (ordered pair - polar form)

\vec F = (30\,lbf, 25^{\circ})

Sum of vectorial components (linear combination)

\vec F = (30\cdot \cos 25^{\circ})\cdot \hat{i} + (30\cdot \sin 25^{\circ})\cdot \hat{j}\,[N]

\vec F = 27.189\cdot \hat{i} + 12.679\cdot \hat{j}\,[N]

8 0
3 years ago
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