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lesantik [10]
3 years ago
6

Algebra 1B: Exponential Functions Unit 2

Mathematics
1 answer:
AfilCa [17]3 years ago
5 0
To be honest. That’s a hard question there
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Look at the number line below.
WARRIOR [948]

B. is the ans hope you get a hundred

5 0
3 years ago
Please help
ElenaW [278]
20 days
x=number of days vacation
y=daily expenses
xy=1100
(x+2)(y-5)=1100
*******
xy-5x+2y-10=1100
xy-5x+2y=1110
y=1100/x
x(1100/x)-5x+2(1100/x)=1110
1100-5x+(2200/x)=1110
-5x+(2200/x)=10
-5x^2 + 2200 = 10x
-x^2 + (440) = 2x
x^2 - 440 + 2x = 0
(x+22)(x-20)=0
x=-22,20
only take the positive answer because the negative answer doesn't make sense in the question
so the answer is 20
****
I might be wrong; this question probation should've been this hard so I might have taken a wrong turn somewhere. feel free to double check my work
3 0
3 years ago
The sum of two numbers is 68 and the difference is 12. What are the numbers? Larger number: Smaller number:
Lady_Fox [76]

Answer:

This is hard asfk

Step-by-step explanation:

Sorry but im confused just as you.

5 0
3 years ago
Simplify the Expression below (2x^2-5x+6)+(5x^2-3x+4)
tia_tia [17]

Answer: 7x^2-8x+10

Step-by-step explanation: hope this helps gl :)

thx for the points

4 0
3 years ago
Read 2 more answers
. All the students in SS3 of a named school take either Mathematics (M), or Physics (P) or Chemistry (C). 40 take Mathematics, 4
aivan3 [116]

Answer:

(a) = 13

(b) = 8

(c) = 5

Step-by-step explanation:

Addition theorems on sets are

Theorem 1 :

n(AuB) = n(A) + n(B) - n(AnB)

Theorem 2 :

n(AuBuC) : = n(A) + n(B) + n(C) - n(AnB) - n(BnC) - n(AnC) + n(AnBnC)

Total number of students in the school is not given

so let there are 60 students in the school

using theorem 2

n(AuBuC) : = n(A) + n(B) + n(C) - n(AnB) - n(BnC) - n(AnC) + n(AnBnC)

let n(A) = Mathematics, n(B) =Physics and n(C) = Chemistry

so putting values,

   60             =    40 + 42 + 38 - 20 - 28 - 25 + n(AnBnC)

60 +73 -120 = n(AnBnC)

         13         = n(AnBnC)

therefore, there are total 13 students who take all three subjects

Number of students who had taken only Mathematics  =
n(A) - n(AnB) - n(AnC) + n(AnBnC)

40 - 20 - 25 + 13

53 - 45 = 8 students

Number of students who had taken only Physics  =

n(B) - n(BnA) - n(BnC) + n(AnBnC)

42 - 20 - 28 + 13

53 - 48 = 5 students

learn more about sets and vein diagrams at

brainly.com/question/2332158

#SPJ10

4 0
2 years ago
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