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Ksju [112]
3 years ago
14

2a2b3 and -4a2b3 are like terms. True False

Mathematics
2 answers:
algol133 years ago
7 0

Answer:


Step-by-step explanation:


AysviL [449]3 years ago
3 0
2a2b3 and -4a2b3 are like terms true 

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7c3d2<br><br> Need help finding the degree of the monomial
Katyanochek1 [597]

Answer:

5

Step-by-step explanation:

7c³d²

This term has tow variables, c and d

Power of c is 3

Power of d is 2

Degree of this monomial is = 3+2 = 5

8 0
3 years ago
PLSS HELP IM CONFUSED ON THIS QUESTION The figure below shows a transversal t which intersects the parallel lines PQ and RS:
n200080 [17]

1. 4 and 8 ==》 <em>corresponding angles</em>

_____________________________

2. 5 and 6 ==》 <em>linear</em><em> </em><em>pair</em><em> </em><em>angles</em>

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3. 2 and 8 ==》 <em>alternate exterior angles </em>

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4. 3 and 6 ==》 <em>same side interior angle</em>s

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5. 2 and 4 ==》 <em>vertical</em><em> </em><em> </em><em>angles</em><em> </em>

8 0
3 years ago
Please help due tmrw
Wewaii [24]

Answer:

1. w = 3l

2. k = 1/2 * b

Step-by-step explanation:

1.  w = 3l

   w = width     l = length

   Since we are solving the equation for the width, we can make the equation w = ? for now. since the width is equal to three times the length, we can make the equation w = 3l.

2. k = 1/2 * b

   k = kool-aid     b = beef    

   Since we are solving the equation for the kool-aid, we can make the equation k = ? for now. since the kool-aid is equal to half the beef sold, we can make the equation k = 1/2 * b

I hope this helped and please mark me as brainliest!

7 0
3 years ago
Read 2 more answers
Emily wants to hang a painting in a gallery. The painting and frame must have an area of 31 square feet. The painting is 5 feet
FromTheMoon [43]

Emily could start with the quadratic equation

... (2x+5)(2x+6)=31

... 4x² +22x +30 = 31

And then subtract the right side to put it into standard form.

... 4x² +22x -1 = 0 . . . . . . corresponds to the first selection

8 0
3 years ago
Read 2 more answers
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65
erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

               Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 2.65

           \sigma  = standard deviation = 0.85

            n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

    P(X bar <= 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z < 2.06) = 0.98030

 P(X bar <= 2.65) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3)  = 0.98030 - 0.5 = 0.4803 .

                                                                             

8 0
4 years ago
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