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4 years ago

What is the coefficient of the xy term in the expression below? 2x + 3y - 4xy - 10

Mathematics

2 answers:

max2010maxim [7]4 years ago

7 0

Answer:

The coefficient of the xy term in the expression is -4

Step-by-step explanation:

In an algebric expression, the coefficient is the number attached to a variable x,y,z.

For example in algebric equation 4x + 3y = 12

The coefficient of x is 4 and the coefficient of y is 3.

So in the given term 2x + 3y - 4xy - 10,

The coefficient of the xy term is -4

alukav5142 [94]4 years ago

3 0

For this case we have the following expression:

$2x + 3y-4xy-10$

By definition, a coefficient is the term that accompanies one or more variables.

For example:

$3x + 5$

The coefficient of the linear term is 3.

So, in the given expression we have to:

-4 is the term that accompanies the variables xy.

Answer:

-4

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Step-by-step explanation:

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3 years ago

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amid [387]

Answer:

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3 years ago

prism x is showen below. the volume of prism y is 10 cubic cm greater than the volume of prism x. what is the volume of prism y.

Vlad1618 [11]

Answer:

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Step-by-step explanation:

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2 years ago

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station. Find the ra

Anika [276]

Answer:

The rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

Step-by-step explanation:

Given information:

A plane flying horizontally at an altitude of "1" mi and a speed of "430" mi/h passes directly over a radar station.

$z=1$

$\frac{dx}{dt}=430$

We need to find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

$y=2$

According to Pythagoras

$hypotenuse^2=base^2+perpendicular^2$

$y^2=x^2+1^2$

$y^2=x^2+1$ .... (1)

Put z=1 and y=2, to find the value of x.

$2^2=x^2+1^2$

$4=x^2+1$

$4-1=x^2$

$3=x^2$

Taking square root both sides.

$\sqrt{3}=x$

Differentiate equation (1) with respect to t.

$2y\frac{dy}{dt}=2x\frac{dx}{dt}+0$

Divide both sides by 2.

$y\frac{dy}{dt}=x\frac{dx}{dt}$

Put $x=\sqrt{3}$ , y=2, $\frac{dx}{dt}=430$ in the above equation.

$2\frac{dy}{dt}=\sqrt{3}(430)$

Divide both sides by 2.

$\frac{dy}{dt}=\frac{\sqrt{3}(430)}{2}$

$\frac{dy}{dt}=372.390923627$

$\frac{dy}{dt}\approx 372$

Therefore the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station is 372 mi/h.

6 0

4 years ago

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ludmilkaskok [199]

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Step-by-step explanation:

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4 0

3 years ago