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slega [8]
3 years ago
6

Guys I need help! I am given two chances to get this question right!

Mathematics
1 answer:
amid [387]3 years ago
8 0

The four digit code is

1563

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What is the volume of this prism? <br> A. 23.9 ^3<br> B. 47.7 cm^3<br> C. 4.50 cm^3<br> D. 9.00 cm^3
melisa1 [442]

Answer: A) 23.9

Step-by-step explanation:

volume of a shape means the area of its base multiplied by its height

so , the base here is a right angle triangle it's area equals 1/2*b*h = 1/2*3*6=9 m² and the prism height is 2.65.

therefore the volume is 2.65*9

8 0
3 years ago
Three out of every 20 students in ms. jones’ Math class earned a grade of A. What percent of the students earned a grade of A?
erica [24]

Answer:

15%

Step-by-step explanation:

3/20x100=15

6 0
2 years ago
PLEASE HELPPPPP. URGEENNTT!!! I DONT KNOW HOW TO SOLVE THIS
Oksi-84 [34.3K]

Answer:

A = 58.7 degrees

B = 66.9 degrees

C = 34.1 degrees

Step-by-step explanation:

<u><em>For <A:</em></u>

Tan A = \frac{opposite}{adjacent}

Tan A = \frac{23}{14}

Tan A = 1.6

A = Tan^{-1} 1.6

A = 58.7 degrees

<u>For <B:</u>

Sin B = \frac{opposite}{hypotenuse}

Sin B = \frac{23}{25}

Sin B = 0.92

B = Sin^{-1} 0.92

B = 66.9 degrees

<em><u>For <C:</u></em>

Sin C = \frac{opposite}{hypotenuse}

Sin C = \frac{14}{25}

Sin C = 0.56

C = Sin^{-1}0.56

C = 34.1 degrees

6 0
3 years ago
HELP PLS I WILL MARK BRAINLYIiEST ASAP!!!!
kicyunya [14]

Answer: 50.84%

Step-by-step explanation:

(89-59) /59 = 0.5084/100 = 50.84%

8 0
3 years ago
given examples of relations that have the following properties 1) relexive in some set A and symmetric but not transitive 2) equ
rodikova [14]

Answer: 1) R = {(a, a), (а,b), (b, a), (b, b), (с, с), (b, с), (с, b)}.

It is clearly not transitive since (a, b) ∈ R and (b, c) ∈ R whilst (a, c) ¢ R. On the other hand, it is reflexive since (x, x) ∈ R for all cases of x: x = a, x = b, and x = c. Likewise, it is symmetric since (а, b) ∈ R and (b, а) ∈ R and (b, с) ∈ R and (c, b) ∈ R.

2) Let S=Z and define R = {(x,y) |x and y have the same parity}

i.e., x and y are either both even or both odd.

The parity relation is an equivalence relation.

a. For any x ∈ Z, x has the same parity as itself, so (x,x) ∈ R.

b. If (x,y) ∈ R, x and y have the same parity, so (y,x) ∈ R.

c. If (x.y) ∈ R, and (y,z) ∈ R, then x and z have the same parity as y, so they have the same parity as each other (if y is odd, both x and z are odd; if y is even, both x and z are even), thus (x,z)∈ R.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial but not transitive, so the relation provided in (1) satisfies this condition.

Step-by-step explanation:

1) By definition,

a) R, a relation in a set X, is reflexive if and only if ∀x∈X, xRx ---> xRx.

That is, x works at the same place of x.

b) R is symmetric if and only if ∀x,y ∈ X, xRy ---> yRx

That is if x works at the same place y, then y works at the same place for x.

c) R is transitive if and only if ∀x,y,z ∈ X, xRy∧yRz ---> xRz

That is, if x works at the same place for y and y works at the same place for z, then x works at the same place for z.

2) An equivalence relation on a set S, is a relation on S which is reflexive, symmetric and transitive.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial and not transitive.

QED!

6 0
3 years ago
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