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3 years ago

What is the area of a rectangle with a length 20 yd and a width of 16 yd?

Mathematics

2 answers:

DochEvi [55]3 years ago

7 0

The answer would be A. 320 yd2 because when you multiply 20x16 you get 320

Jlenok [28]3 years ago

6 0

A. 320 yd2 16yd times 20yd

You might be interested in

B is between A and C. AB = 3x, BC = 2x + 7, AC = 42 a) X = b)AB = c)BC=

ryzh [129]

Answer: x=7

ab=21

bc=14

Step-by-step explanation:

ab=3x

bc=2x±7

ac=42

3x±2x±7=42

5x±7=42

5x=42-7

5x=35

to get the value of x you divide 35 by 5x

35÷5x=7

so the value of x is 7

ab=3x

x=7

3x=7×3x

7×3=21

x=7

2x=7×2x

2×7=14

5 0

2 years ago

Solve the following equation fro g:2(g-h)=b+4

monitta

2g - 2h = b + 4
2g = b + 4 + 2h
g = (b + 4 + 2h) / 2

4 0

2 years ago

Read 2 more answers

Simplify. 2^5−4⋅3^3 7+(1+100√)

frez [133]

$\\ \sf\bull\longmapsto 2^5-4(3^3)$

$\\ \sf\bull\longmapsto 32-4(27)$

$\\ \sf\bull\longmapsto 32-108$

$\\ \sf\bull\longmapsto -76$

7 0

2 years ago

Six people are standing in line for concert tickets. Their average age is 27. The average age of the first four people in line i

scZoUnD [109]

Let the age of the 4th person be F
Sum of the 1st four in line: 4(23), or 92
Sum of 1st 3: 92 - F (excludes the 4th person's age)
Sum of the last 3 in line: 3(34), or 102
Sum of last 2: 102 - F (excludes the 4th person's age)
Sum of all 6: 6(27), or 162 (INCLUDES 4th person's age)
We now get: F + 92 - F + 102- F = 162
- F + 194 = 162
- F = 162 - 194
- F = - 32
F, or 4th person's age = -32/-1, or 32

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6 0

3 years ago

There are 2504 computer science students at a school. Of these, 1876 have taken a course in Java, 999 have taken a course in Lin

Y_Kistochka [10]

Answer:

492

Step-by-step explanation:

As per the given data of the question:

Total number of students = 2504

Number of students in Java (J) = 1876

Number of students in Linux (L) = 999

Number of students in C = 345

J∩L = 876

L∩C = 231

C∩J = 290

L∩J∩C = 189

Now according to Venn-diagram as drawn below:

Number of students haven taken courses in Java and Linux both only

= J∩L - L∩J∩C

= 876 - 189

= 687

Number of students haven taken courses in Java and C both only

= C∩J - L∩J∩C

= 290 - 189

= 101

Number of students haven taken courses in C and Linux both only

= L∩C - L∩J∩C

= 231 - 189

= 42

Therefore,

Number of students only in Java = 1876 - 687 - 189 - 101 = 899

Number of students only in Linux = 999 - 687 - 189 - 42 = 81

Number of students only in C = 345 - 42 - 189 - 101 = 13

So,

Number of students who have not taken a course in any of these three subjects

= 2504 - 899 - 81 - 13 - 687 - 189 -101 - 42

= 492

Hence, the students who have not taken a course in any of these three subjects = 492.

6 0

3 years ago