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ziro4ka [17]
3 years ago
12

PLZ HELP!!! I will give you brainliest

Mathematics
1 answer:
Artist 52 [7]3 years ago
6 0

Answer:

141

Step-by-step explanation:

CD = | 4-(-2)| = 6

DB = |3-(-7)| = 10

area of a right triangle

A = (leg 1 x leg2)/2 = (6 x 10)/2 = 30

EB = |3-(-3)| = 6

EF = |-8-(-2)| = 6

area of a square = side^2

A = 6^2 = 36

segment G(and a point with same x coordinate of H)

|8-3| = 5

segment H(and the same point)

|-10-(-8)| = 2

A = (5x2)/2 = 5

segment I(and the same point)

|-8-6| = 14

Area of a rectangle

A = base x width

A = (5 x 14) = 70

total area = 70 + 5 + 36 + 30 = 141

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Lina20 [59]

Step-by-step explanation:

mình nghĩ là như vầy. Chúc bạn học tôt :))))

8 0
3 years ago
Wich numbers are rational (r) or irrational (i)? 68,18/42,70.2329,91.154389,5.
skelet666 [1.2K]
68 is rational
18/42 is rational
70.2.29 is rational
91.154389 is rational
5 is 

3 0
3 years ago
Read 2 more answers
3,0 and 4,l what is the y-intercept of the line
alexandr1967 [171]

Answer:

Step-by-step explanation:

(3,0),(4,1) ⇒ (x₁,y₁),(x₂,y₂)

to find the equation of the line

y-y₁= {(y₂-y₁)/(x₂-x₁)}(x-x₁)

y-0= [(1-0)/(4-3)] (x-3)

y-0 = 1/1(x-3)

y-0 = x-3

y= x-3+0

y=x-3

compare to y=mx+c, where m is the gradient and c is the intercept on y-axis

m=1, c=-3

∴the y-intercept of the line is -3

7 0
3 years ago
Find the sum of the series and show your process for finding each term leading to the final sum.
Crazy boy [7]

ANSWER

\sum_{k=3}^5( - 2k + 5) =  - 9

EXPLANATION

The given series is

\sum_{k=3}^5( - 2k + 5)

This series is finite.

The expanded form is

\sum_{k=3}^5( - 2k + 5) = ( - 2 \times 3 + 5) + ( - 2 \times 4 + 5) + ( - 2 \times 5 + 5)

This simplifies to

\sum_{k=3}^5( - 2k + 5) = ( - 6 + 5) + ( - 8+ 5) + ( - 10+ 5)

\sum_{k=3}^5( - 2k + 5) = ( - 1) + ( - 3) + ( - 5)

\sum_{k=3}^5( - 2k + 5) =  - 9

5 0
2 years ago
PLEASE HELPPPP Prove That: 5^31–52^9 is divisible by 100. do it like _____*100
Andru [333]

Answer:

5^31

Step-by-step explanation:

10052^9

This should be correct

8 0
3 years ago
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