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marta [7]
3 years ago
12

Please help! will mark brainliest

Mathematics
1 answer:
marin [14]3 years ago
7 0

The correct answer is D.

x= -4 and y=2

left 4  up 2


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Please answer the file question.
viktelen [127]

Answer:48 16 = ?

27 9 = 9

60 20 - 10

Step-by-step explanation:

4 0
2 years ago
Round your answer to the nearest tenth.
Kruka [31]

Answer:

The value of MB = 8.4

Step-by-step explanation:

We know that the point of intersection of the Medians of a triangle is called the centroid of a triangle.

Thus,

For the given triangle ΔJKL,

  • The point M is the centroid of the triangle.

We also know that the centroid is 2/3 of the distance from each vertex to the midpoint of the opposite side.

Also, each Median is split into two parts such that the longer part is 2 times the length of the smaller part.

In our case,

The median KB is split into two parts such that the longer part KM is 2 times the length of the smaller part MB.

i.e.

KM = 2 MB

Given KM = 16.8

so substitute KM = 16.8 in the equation KM = 2 MB

16.8 = 2 MB

MB = 16.8/2

MB = 8.4

Therefore, the value of MB = 8.4

7 0
3 years ago
Read 2 more answers
I need help with this question asap!!
Sholpan [36]

Answer:

1/4= 25%

Step-by-step explanation:

6 0
2 years ago
Help me pleaseeeessss
wlad13 [49]

Answer:

28/53

Step-by-step explanation:

Hey There!

So they want us to find the sine of angle S

well remember is sohcahtoa

S - sine

O - opposite

H - hypotenuse

meaning that

Sin=\frac{opposite}{hypotenuse}

The opposite side of angle S is 28 and the hypotenuse is 53

so sinS=\frac{28}{53}

5 0
2 years ago
You flip a coin and then roll a 6-sided number cube (a die).
expeople1 [14]

Answer:

a)  No, it does not matter whether you roll the die or flip the coin first, as these two events are <u>independent</u> of each other, which means they do not affect each other.

b) Yes.

  • Let event 1 be flipping a coin and event 2 be rolling a die.
  • Let event 1 be rolling a die and event 2 be flipping a coin.

The likelihood that any outcome will occur will not change, as the events are independent.

c) see attached

d)   12 outcomes  (H = head, T = tail, numbers represent the value of the die)

H 1           T 1

H 2          T 2

H 3          T 3

H 4          T 4

H 5          T 5

H 6          T 6

e)  

\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}

\implies \sf P(even)=\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}

\implies \sf P(head)=\dfrac{1}{2}

\implies \sf P(even)\:and\:P(head)=\dfrac{1}{2} \times \dfrac{1}{2}=\dfrac{1}{4}

6 0
2 years ago
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