MATHEMATICS HIGH SCHOOL

What is the correct classification of 18x?

Answers

Answer 1

Answer: monomial!!!!!!!!!!!!!

Answer 2

Answer:

Answer:

monomial

Step-by-step explanation:

Related Questions

Ok what is the answer
Term insurance has value as an investment. True or false ?
25d-16=4d+26 whats d?
What error did she make? She simplified the denominator incorrectly. The denominator simplifies to –7. She labeled the points incorrectly. The point (–7, 4) should be (x1, y1). She used an incorrect formula. The formula should be the change in y-values with respect to the change in the x-values. She used an incorrect formula. The formula should be the sum of the x-values with respect to the sum of the y-values.
70 ans 20 in simplest fore in ratios

the perimeter of a square is 56 cm. what is the approximate length of its diagonal? 10.6 cm 14.0 cm 15.0 cm 19.8 cm

Answers

we know that

$Perimeter\ of\ a\ square=4b$

where

b is the length side of a square

Step $1$

Find the length side b of a square

$P=56cm\n P=4b\n 56=4b\n b=56/4\n b=14cm$

Step $2$

Find the length of the diagonal of a square

Applying the Pythagorean Theorem

$D=\sqrt{b^(2)+b^(2)} \n D=\sqrt{14^(2)+14^(2)}\n D=14√(2) cm\n D=19.80cm$

therefore

the answer is

$19.80cm$

The answer to this problem is 19.8

A crane able can support a maximum load of 15,000 kg. If a bucket has a mass of2,000 kg and gravel has a mass of 1,500 kg for every cubic meter, how many cubicmeters of gravel can be safely lifted by the crane?

Answers

Let

x-------> amount of cubic meters of gravel that can be safely lifted by the crane

we know that

$15,000=2,000+1,500*x$

Solve for x

$15,000=2,000+1,500*x\n1,500*x=15,000-2,000\n1,500*x=13,000\n x=(13,000/1,500)\nx=8.67\ m^(3)$

therefore

the answer is

the amount of cubic meters of gravel that can be lifted safely by the crane cannot exceed $8.67\ m^(3)$

Well so 15000-2000 leaves 13000kgs of available weight allowed. Then 13000 divided by 1500 we get around 8.67 cubicmeters of gravel to be allowed to be lifted safely.

The equation represents an ellipse. which points are the vertices of the ellipse? (−11, 0) and (11, 0) (−1, 0) and (1, 0) (0, −11) and (0, 11) (0, −1) and (0, 1)

Answers

The answer is A (-11,0) & (11,0)

Answer:

(−11, 0) and (11, 0)

Step-by-step explanation:

A wildflower bouquet costs $4.25 at the farmer's market. A vase can be purchased for an additional $3.50. Kelly has $20 to buy a bouquet for her mother with a vase. She'd also like to buy just a bouquet for herself and another for her best friend.If x represents the number of bouquets Kelly can buy, which inequality will tell Kelly if she has enough money to make her purchase?
A) $4.25x + $3.50 ≤ $20
B) $4.25x - $3.50 ≤ $20
C) $4.25x + $3.50 ≥ $20
Eliminate
D) $3.50x + $4.25 ≤ $20;

Answers

A is the answer.
4.25 multiply 3= 12.75
12.75 + 3.50= 16.50
16.50 ≤ $20

Answer:

Step-by-step explanation:

Jonathan has been on a diet since January 2013. So far, he has been losing weight at a steady rate. Based on monthly weigh-ins, his weight, w, can be modeled by the function w=-3m+205 where m is the number of months after January 2013a) How much did Jonathan weigh at the start of the diet?

b) How much weight has Jonathan been losing each month?

c) How many month did it take Jonathan to lose 45 pounds?

Show work plz

Answers

$a)\ \ \ w(m)=-3m+205\n\nm=0\ \ \ - at\ the\ start\ of\ the\ diet\n \nw(0)=-3\cdot0+205=205\n \nAns.\ weigh\ at\ the\ start\ of\ the\ diet\ is\ 205\ pounds\n \nb)\ \ \ w(m+1)=-3(m+1)+2015=-3m-3+205=-3m+202\n \nw(m+1)-w(m)=-3m+202-(-3m+205)=\n \n=-3m+202+3m-205=-3\n \nAns.\ Jonathan\ has\ each\ month\ 3\ been\ losing \n \n w(m)=205-45=160\ \ \ \Rightarrow\ \ \ -3m+205=160\ \ \ \Rightarrow\ \ \-3m=-45\n \nm=15\n \nAns.\ 15\ month$

The daily high temperatures, in degrees Celsius, for two weeks were recorded. The first week had a mean high temperature of 8 degrees Celsius and a mean absolute deviation (MAD) of 1 degree Celsius. The second week had a mean high temperature of 4.86 degrees Celsius and a mean absolute deviation (MAD) of 1.91 degrees Celsius. Which week had greater variability in high temperature? Explain your reasoning.

Answers

The required week that had greater variability in high temperature is week 2.

Given that,
The first week had a mean high temperature of 8 degrees Celsius and a mean absolute deviation (MAD) of 1 degree Celsius.
The second week had a mean high temperature of 4.86 degrees Celsius and a mean absolute deviation (MAD) of 1.91 degrees Celsius.

What is Statistic?

Statistics is the study of mathematics that deals with relations between comprehensive data.

Here,
The system of observation that has a higher mean absolute deviation, will have the chance of greater variability in high temperature. So,
mean absolute deviation of week 1 = 1
mean absolute deviation of week 2 = 1.91
MAD of week 2 > MAD of week 1

Thus, the required week that had greater variability in high temperature is week 2.

Learn more about Statistics here:
brainly.com/question/23091366

#SPJ2

Answer:

mean 4.86

MAD = 1.91

Step-by-step explanation:

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