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PtichkaEL [24]
3 years ago
13

Fill in the blanks with '<'or'>?.

Mathematics
1 answer:
Sveta_85 [38]3 years ago
5 0

Answer:

See below.

Step-by-step explanation:

(a) 0.4 m < 0.6m

(b) 0°C > -4°C

(c) -2°C > -2.5°C

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What is the first operation used to evaluate 13-2x3+4 divided by 4+4
lisabon 2012 [21]

Answer:

  multiplication

Step-by-step explanation:

The evaluation of ...

  13 -2·3 +4/4 +4

starts with the multiplication, because there are no exponents or parentheses.

  13 -6 +4/4 +4

Next is the division:

  13 -6 +1 +4

Finally, the addition and subtraction:

  7 +1 +4

  8 +4

 12

_____

We have assumed your x is not a variable, but is intended to indicate multiplication. We have also assumed that your "divided by" implies no particular grouping, so that the numerator is the first preceding number and only the first following number is in the denominator.

6 0
2 years ago
Domain and range???!!!!!!
elena55 [62]

Answer:

domain={1,-3}

range={-5,1,2,4}

6 0
2 years ago
The coordinate of centroid of a triangle whose vertices are (1,3,-2), (4,5,0), (6,3,9) is = ……………….
Dominik [7]

Answer:

C = (\frac{11}{3},\frac{11}{3},\frac{7}{3})

Step-by-step explanation:

Given

(x_1,y_1,z_1) = (1,3,-2)

(x_2,y_2,z_2) = (4,5,0)

(x_3,y_3,z_3) = (6,3,9)

Required

Determine the coordinates of the centroid

Represent the coordinates with C.

C is calculated as follows:

C = (\frac{1}{3}(x_1+x_2+x_3),\frac{1}{3}(y_1+y_2+y_3),\frac{1}{3}(z_1+z_2+z_3}))

Substitute values of x and y in the given equation

C = (\frac{1}{3}(1+4+6),\frac{1}{3}(3+5+3),\frac{1}{3}(-2+0+9}))

C = (\frac{1}{3}(11),\frac{1}{3}(11),\frac{1}{3}(7}))

C = (\frac{11}{3},\frac{11}{3},\frac{7}{3})

<em>The above is the coordinate of the centroid</em>

8 0
2 years ago
"Eight more than twice a number" can be written as what algebraic expression?
Ainat [17]
" 8 more "....means add 8
" twice a number " means multiply the number by 2

ur expression is : 2n + 8
7 0
3 years ago
Read 2 more answers
At a recent marathon, spectators lined the street near the starting line to cheer for the runners. The crowd lined up 5 feet dee
masha68 [24]

Answer:

29568 people cheered for the runners at the start of the race

Step-by-step explanation:

From the question, the crowd lined up 5 feet deep on both sides of the street for the first mile.

This lined up crowd could be related to a rectangle that is 1 mile long and 5 feet wide.

First, Convert 1 mile to feet

1 mile = 5280 feet

Hence, the length of the rectangle is 5280 feet and the width is 5 feet.

Now, we will determine how many 5 feet by 5 feet square we can get from the 5280 feet by 5 feet rectangle. To do that, we will divide 5280 feet by 5 feet

5280 feet ÷ 5 feet = 1056

Hence, from the rectangle, we can get 1056 5 feet by 5 feet square.

From the question, you estimate that 14 people can comfortably fit in a square that measures 5 feet by 5 feet,

∴ 14 × 1056 people will comfortably fit in the crowed lined up 5 feet deep on one side of the street for the first mile.

14 × 1056 = 14784 people

This is the amount of people that will comfortably fit in the crowed lined up 5 feet deep on one side of the street for the first mile.

Since the crowd lined up on both sides of the street, then

2 × 14784 people will comfortably fit in the crowed lined up 5 feet deep on both sides of the street for the first mile

2 × 14784 = 29568 people

Hence, 29568 people cheered for the runners at the start of the race.

5 0
3 years ago
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