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Leokris [45]
2 years ago
12

30 POINTS NEED HELP ASAP MATH QUESTION A. 19 B. 108 C. 133 D. 763

Mathematics
1 answer:
Oduvanchick [21]2 years ago
3 0

Answer:

lol that's 14 points my boy

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✓ 10
liraira [26]

Answer:

Decrease in dollars is $555.00, $6845.00 was in his account at the end of last year.

Step-by-step explanation:

7400 times 7.5/100

=74 times 7.5

=$555.00

7400-555=$6845.00

7 0
2 years ago
A solid shape is made by joining two cones.
dlinn [17]

The surface area of the solid shape is the amount of space on it

The radius of the two cones is 2.93 cm

<h3>How to determine the radius?</h3>

The given parameters are:

  • Cone 1: slant height, l = 2r
  • Cone 2: slant height, L = 3r
  • Surface area = 135. 21 cm²

The surface area of the shape is calculated using:

T = πr(L + l)

So, we have:

135.21 = πr(2r + 3r)

Evaluate

135.21 = 5πr²

Divide both sides by 5π

r² = 135.21/5π

Evaluate the quotient

r² = 8.61

Take the square root of both sides

r = 2.93

Hence, the radius of the cones is 2.93 cm

Read more about surface area at:

brainly.com/question/6613758

3 0
2 years ago
Emily read 24 fiction books and 2 nonfiction books. What is the ratio of the number of fiction books she read to the total numbe
dimulka [17.4K]

Answer:

26

Step-by-step explanation:i used a caluculator

3 0
2 years ago
Read 2 more answers
Check whether the relation R on the set S = {1, 2, 3} is an equivalent
kozerog [31]

Answer:

R isn't an equivalence relation. It is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

Let S denote a set of elements. S \times S would denote the set of all ordered pairs of elements of S\!.

For example, with S = \lbrace 1,\, 2,\, 3 \rbrace, (3,\, 2) and (2,\, 3) are both members of S \times S. However, (3,\, 2) \ne (2,\, 3) because the pairs are ordered.

A relation R on S\! is a subset of S \times S. For any two elementsa,\, b \in S, a \sim b if and only if the ordered pair (a,\, b) is in R\!.

 

A relation R on set S is an equivalence relation if it satisfies the following:

  • Reflexivity: for any a \in S, the relation R needs to ensure that a \sim a (that is: (a,\, a) \in R.)
  • Symmetry: for any a,\, b \in S, a \sim b if and only if b \sim a. In other words, either both (a,\, b) and (b,\, a) are in R, or neither is in R\!.
  • Transitivity: for any a,\, b,\, c \in S, if a \sim b and b \sim c, then a \sim c. In other words, if (a,\, b) and (b,\, c) are both in R, then (a,\, c) also needs to be in R\!.

The relation R (on S = \lbrace 1,\, 2,\, 3 \rbrace) in this question is indeed reflexive. (1,\, 1), (2,\, 2), and (3,\, 3) (one pair for each element of S) are all elements of R\!.

R isn't symmetric. (2,\, 3) \in R but (3,\, 2) \not \in R (the pairs in \! R are all ordered.) In other words, 3 isn't equivalent to 2 under R\! even though 2 \sim 3.

Neither is R transitive. (3,\, 1) \in R and (1,\, 2) \in R. However, (3,\, 2) \not \in R. In other words, under relation R\!, 3 \sim 1 and 1 \sim 2 does not imply 3 \sim 2.

3 0
3 years ago
I need help plz will gave brainliest​
nata0808 [166]

m1 = 63 m2 =117 m3=117 m5=117 m6=63 m7=117 m8=63

7 0
2 years ago
Read 2 more answers
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