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3 years ago

How do you recognize when an equation has like terms

1 answer:

galina1969 [7]3 years ago

7 0

Like terms" refers to individual terms that have the same variable in them. Also, the variable in any "like terms" is raised to the same power. For example: 3x and 10x are like terms (both have the variable x, raised to the 1st power). 14a2 and ba2 are also like terms (both have the variable a, raised to the 2nd power)

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15 decreased by twice a number x in algebriac expression

postnew [5]

2x - 15

15 is decreased that is the key word here which indicates subtraction.

twice a number with given variable x is 2x

subtraction always goes behind so 2x - 15

4 0

3 years ago

Which expression represent a number that is one fourth as great as 10-2

Nataliya [291]

I think the answer to the expression would be D. (10 - 2) x 4 because (10 - 2) would be multiplied 4 times, which means it would be four times greater. I hope this made sense and also helped you in some way.

5 0

3 years ago

Read 2 more answers

Simplify x^6+10x^5+25x^4-x^2-10x-25 using synthetic division

antiseptic1488 [7]

Saan po yung divisor?

Kailangan ko po

7 0

2 years ago

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks along the x-axis from the spotlight toward the bui

Sedbober [7]

Answer:

0.675 m/s

Step-by-step explanation:

Let height of shadow= y,CD=x

Height of man=2 m

Speed of man= $\frac{dx}{dt}=1. 8 m/s$

$\triangle ABD\sim\triangle ECD$

Therefore, $\frac{AB}{EC}=\frac{BD}{CD}$

$\frac{y}{2}=\frac{12}{x}$

$xy=24$

Differentiate w.r.t t

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

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

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

When the man is 4 m from the building

Then, we have x=12-4=8 m

$\frac{dx}{dt}=1.8 m/s$

Substitute the values in above equation then, we get

$8y=24$

$y=\frac{24}{8}=3$

Substitute the values then we get

$\frac{dy}{dt}=-\frac{3}{8}\times 1.8=-0.675 m/s$

Hence, the length of his shadow on the building decreasing at the rate 0.675 m/s.

8 0

3 years ago

Create a set of coordinates to model a relation that is a linear function.

STALIN [3.7K]

I will create a set of arbitrary constants (x1,y1) (x2,y2)

slope = y2-y1/x2-x1

y = (y2-y1/x2-x1)x + b

y2 = (y2-y1/x2-x1)x2 + b

b = y2 - (y2-y1/x2-x1)x2

y = (y2-y1/x2-x1)x + [y2 - (y2-y1/x2-x1)]

Choose any points and just
Plug the values and you have a linear function.

NOT SURE IF THAT'S WHAT THE QUESTION WANTS.

3 0

2 years ago