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trasher [3.6K]
3 years ago
10

How many terms does the following linear expression have?

Mathematics
1 answer:
sertanlavr [38]3 years ago
3 0

Answer:

4 terms.

Step-by-step explanation:

x^2 has 2 terms/

3x is one term.

1/5x is one term.

Basically the number of x's.

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Ksenya-84 [330]
The answer is G. Lucy
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Suppose x=18,y=24, and z=30 in the figure, which is not drawn to scale. Determine the following values. Enter exact answers (as
sammy [17]

Answer:

See explanation

Step-by-step explanation:

cos(θ)= 18/30 = 3/5

sin(θ)= 24/30 = 4/5

tan(θ)= 24/18 = 4/3

cos(ϕ)= 24/30 = 4/5

sin(ϕ)= 18/30 = 3/5

tan(ϕ)= 18/24 = 3/4

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How do you do 185 times 12 in standard algorithm?
jolli1 [7]

the answer is 2,220

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6 0
3 years ago
Mr. Mole left his burrow and started digging his way down at a constant rate.
Art [367]

Mr. Mole's burrow was at an altitude of 6 meters below the ground.

Step-by-step explanation:

Step 1:

We need to determine the distance that Mr. Mole covers in a single minute.

To do that we divide the difference in values of altitude by the difference in the time periods.

For the first case, Mr. Mole had traveled -18 meters in 5 minutes.

We also have, he traveled -25.2 meters in 8 minutes.

Step 2:

The distance he covered in 1 minute = \frac{-25.2-(-18)}{8-5} = \frac{-25.2+18}{3},

\frac{-25.2+18}{3} = \frac{-7.2}{3} = -2.4.

So with every minute, Mr. Mole digs down an additional 2.4 meters below the surface.

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So Mr. Mole's burrow was at an altitude of 6 meters below the ground i.e. -6 meters.

5 0
3 years ago
Read 2 more answers
How do I factor out a constant before factoring out a quadratic? <br>Factor completely: 3w^2-3w-90
pantera1 [17]
3w^2-3w-90 =0 \ \ \:3 \\ \\w^2- w-30 =0 \\ \\a=1 , \ b= -1 , c= -30 \\ \\\Delta =b^2-4ac = (-1)^2 -4\cdot1\cdot (-30) =1+120=121 \\ \\w_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{1-\sqrt{121}}{2 }=\frac{ 1-11}{2}=\frac{-10}{2}=-5

w_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{1+\sqrt{121}}{2 }=\frac{ 1+11}{2}=\frac{12}{2}=6 \\ \\ Answer:\\ \\ w^2- w-30 =(x+5)(x-6)
 

3 0
3 years ago
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