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SashulF [63]
3 years ago
15

Can someone plz solve this for me??

Mathematics
2 answers:
erma4kov [3.2K]3 years ago
7 0
I think it’s 67 but I’m not sure
Tamiku [17]3 years ago
4 0

Answer:67



Step-by-step explanation:


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16 There are 90 girls and 60 boys in the sixth grade at a middle school. Of these students, 9 girls
muminat

Answer:

B

Step-by-step explanation:

7 0
4 years ago
Fie O centrul unui cerc si doua unghiuri la centru adiacente AOB si BOC.Stiind ca masura unghiului AOB = 120 si masura BOC= 150,
kodGreya [7K]

<span>Let A be the center of a circle and two angles at the adjacent center AOB and BOC. Knowing the measure of the angle AOB = 120 and the measure BOC = 150, find the measures of the angles of the ABC triangle.
</span>solution

Given the above information;
AC=AB, therefore ABC is an isosceles triangle. 
therefore, BAO=ABO=(180-120)/2=30
OAC=OCA=(180-90)/2=45
OBC=BCO=(180-150)/2=15
THUS;
BAC=BAO+OAC=45+30=75
ABC=OBA+CBO=15+30=45
ACB=ACO+BCO=15+45=60


4 0
3 years ago
Help me out please its with triangles and stuff
OverLord2011 [107]

Answer:

11 in = 33 in.

Step-by-step explanation:

6 in. = 18 in.

the second triangle is 3 times bigger than the smaller one.

7 0
3 years ago
Read 2 more answers
What is the equation in slope-intercept form for the line that passes
igomit [66]

Answer:

y = 6x + 9

Step-by-step explanation:

Δy =-3-3 = -6

Δx =-2-(-1) = -1

Slope = Δy/Δx = 6

Point-slope equation for line of slope 6 that passes through (-1,3):

y-3 = 6(x+1)

Rearrange to solve for y:

y = 6x + 9

5 0
3 years ago
What are the orders of 3,7,9,11,13,17 and 19(mod20)?does 20 have primitive roots?
bezimeni [28]
3\equiv3\mod{20}
3^2\equiv9\mod{20}
3^3\equiv27\equiv7\mod{20}
3^4\equiv3\cdot3^3\equiv3\cdot7\equiv21\equiv1\mod{20}

7\equiv7\mod{20}
7^2\equiv49\equiv9\mod{20}
7^3\equiv7\cdot7^2\equiv63\equiv3\mod{20}
7^4\equiv7\cdot7^3\equiv21\equiv1\mod{20}

9\equiv9\mod{20}
9^2\equiv3^4\equiv1\mod{20}

11\equiv11\mod{20}
11^2\equiv121\equiv1\mod{20}

13\equiv-7\equiv13\mod{20}
13^2\equiv169\equiv9\mod{20}
13^3\equiv13\cdot13^2\equiv(-7)9\equiv-63\equiv-3\mod{20}
13^4\equiv13\cdot13^3\equiv(-7)(-3)\equiv21\equiv1\mod{20}

17\equiv-3\equiv17\mod{20}
17^2\equiv(-3)^2\equiv9\mod{20}
17^3\equiv(-3)^3\equiv-27\equiv3\mod{20}
17^4\equiv(-3)^4\equiv81\equiv1\mod{20}

19\equiv-1\equiv19\mod{20}
19^2\equiv19(-1)\equiv-19\equiv1\mod{20}

Generally speaking, a number x coprime to n will be a primitive root of n if we have x^n\equiv x\mod{n}, or x^{n-1}\equiv1\mod{n}. In other words, if x is of order n-1 modulo n, then x is a primitive root of n.

Since none of these numbers has order 19, it follows that 20 does not have any primitive roots.
6 0
4 years ago
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