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Dovator [93]
3 years ago
14

Can someone explain how to find the missing side lengths.? thank you.

Mathematics
1 answer:
Kay [80]3 years ago
5 0
Use Pythagorean Therom since the triangle is a right triangle. a^2 + b^2 = c^2
a- side
b- side
c) hypotenuse
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Hi! ❤️ I’m looking for help here thanks.
abruzzese [7]

Answer:

i dont remember how to do this and i just did this last year

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Suppose that 1 kg of Sodium-24 is reduced to 0.0156 kg in 90 hours. What is the half-life, in hours, of Sodium-24?​
N76 [4]

Answer:

14.99 hours

Step-by-step explanation:

The formula for half-life is given as,

R/R' = 2ᵃ/ᵇ...................... Equation 1

Where R= Original mass of sodium-24, R' = mass of Sodium-24 after disintegration, a = Total time taken for sodium-24 to disintegrate, b = half-life of sodiun-24.

Given: R = 1 kg, R' = 0.0156 kg, a = 90 hours

Substitute into equation 1 and solve for b

1/(0.0156) = 2⁹⁰/ᵇ

Taking the log of both side

log[1/(0.0156)] = log(2⁹⁰/ᵇ)

log[1/(0.0156)] = 90/b(log2)

90/b = log[1/(0.0156)]/log2

90/b = 6.0023

b = 90/6.0023

b =  14.99 hours

b = 14.99 hours.

Hence the half-life of Sodium-24 is 14.99 hours

7 0
3 years ago
Simplify by combining like terms -17+2(6x-1)+5
earnstyle [38]

\huge\boxed{12x-14}

Start with the original expression.

-17+2(6x-1)+5

Use the Distributive Property — multiply 2 by both 6x and -1.

-17+12x-2+5

Combine the like terms — add up all the constants.

-14+12x

Reorder the terms.

\large\boxed{12x-14}

5 0
3 years ago
Read 2 more answers
Please help ASAP ! Will give brainliest!
Svet_ta [14]

27) x>-6

28) x <7

29) x<= -6

30) x < 7

31) x>=-6

32) x>4

33) x > 8

34) x < 4

35 x >= 5

5 0
3 years ago
The sum of three numbers is 10. Two times the second number minus the first number is equal to 12.
eimsori [14]

<u>Answer:</u>

-10,1,19

<u>Step-by-step explanation:</u>

<u></u>

x+y+z = 10 (Equation 1)

2y-x= 12  (Equation 2)

x-y+2z = 7 (Equation 3)

(Equation 2): -x = -2y+12

                       x = 2y-12 (Equation 4)

(Equation 1) - (Equation 3): 2y-z = 3

                                              -z = -2y+3

                                               z = 2y-3 (Equation 5)

Substitute (4) and (5) into (1)

x+y+z = 10

(2y-12)+y+(2y-3) = 10

5y-15 = 10

5y = 5

y=1

Substitute y=1 into (2)

2y-x= 12

2(1)-x= 12

2-x= 12

-x= 12-2

-x= 10

x= -10

Substitute y=1 and x=-10 into (1)

x+y+z = 10

-10+1+z = 10

z-9 = 10

z = 10+9

z = 19

Order: x = -10, y = 1, and z = 19

8 0
3 years ago
Read 2 more answers
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