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TEA [102]
3 years ago
11

17 subtracted from x is at least 19

Mathematics
1 answer:
Rzqust [24]3 years ago
8 0
X=36.... because 17-36=19
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Rewrite the equation 6x + 7y = 14 into slope-intercept form (y = mx + b)
Tamiku [17]

rearrange into the form y = mx + b

subtract 6x from both sides of the equation

7y = - 6x + 14

divide all terms by 7

y = - \frac{6}{7} x + 2 → in slope- intercept form


6 0
3 years ago
Read 2 more answers
Does the equation x2 - 6x + y2 - 6y = -9 intersect the y-axis
Lana71 [14]
The y intercept is= (0,3)
6 0
3 years ago
Based only on the information given in the diagram, which congruence
professor190 [17]

The congruence theorems or postulates that could be given as reasons why ACDE=AOPQ are:

  • AAS
  • HA
  • ASA

<h3>What is this  congruence theorems about?</h3>

Note that in the diagram, there has been given one side that is congruent to its other corresponding side and as such, it has remove the option or postulates that needs the HL and SAS sides.

Conclusively, when all three angles are said to be congruent and when they have one side (known as the hypotenuse), one can use the congruence theorems or postulates of  AAS, HA and ASA.

Learn more about congruence theorems from

brainly.com/question/10609346

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8 0
2 years ago
Bethany sells roses and petunias. The expression 3r+2.5p3r+2.5p3, r, plus, 2, point, 5, p gives the cost (in dollars) of rrr ros
Lelechka [254]

Answer: The cost of 7 roses and 8 petunias is $41.

Step-by-step explanation:

Given, Bethany sells roses and petunias. The expression 3r+2.5p, gives the cost (in dollars) of r roses and p petunias.

Now  to find the cost of 7 roses and 8 petunias , put r= 7 and p= 8 in the above expression , we get

3(7)+2.5(8)=21+20\\\\=41

Hence, the cost of 7 roses and 8 petunias is $41.

5 0
3 years ago
Couple has got 5 children , what is the probability of have 3girls and 2 boys ?
Delicious77 [7]
5 children so you have 2^5=32 possibilities to "assign" genders
P(3 girls):
how many possibilities are there to "assign" the 3 girl-genders to the 5 children? the first girl has 5 possibilities then the next 4, 3 -> 5*4*3=60
but these possibilities include orders of assigned genders, while children 1-5 might differ the gender "girl" is always the same so we have the remove the orderings of the 3 girl-gender assignments which is 3*2*1=6
if we divide 60/6 we get 10 possibilities to have 3 girls, so what is the resulting chance? the 10 possibilities divided by the total 32 possibilities: 10/32=5/16=P(3 girls)=P(2 boys)

this is a bit of lengthy way of saying "use the binomial coefficient" equation/explaining it a bit which is (n!)/(k!(n-k)!) with n=5, k=3:
5*4*3*2*1/((3*2*1)*(2*1))=
5*4*3*2/(3*2*2)=
5*4*3*2/(3*4)=
5*2=
10 possibilities again


P(girls>=4)=P(boys<=1)=P(boys=1)+P(boys=0)
(or P(girls=4)+P(girls=5))

P(boys=0) is the easy case: simply multiply the chance of getting a girl 5 times: (1/2)^5=1/32
P(boys=1)= again the binomial coefficient with n=5 and k=1:
5*4*3*2*1/((1)*(4*3*2*1))=
5*4*3*2/(4*3*2)=
5 possibilities
so the P(boys=0)=1 possibility + P(boys=1)=5 possibilities totals to 6 possibilities
again the chance is the 6 possibilities divided by all 32 possibilities: 6/32=3/16

P(alternate gender starting with boy): when thinking about the possibilities then there is only a single way to build that order: bgbgb, so one possibility
knowing there is only one way we already know P(alternate...)=1/32 by again dividing by the total amount of possibilities
the alternative way would be to multiply P(boy)*P(girl)*P(boy)*P(girl)*P(boy)=(1/2)^5= 1/32 again





5 0
3 years ago
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