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3 years ago

Tim is 5 years older than JoAnne. Six years from now the sum of their ages will be 79. How old are they now???

Mathematics

2 answers:

alexandr1967 [171]3 years ago

5 0

Tim would be 75 and JOAnne would be 70 years

nasty-shy [4]3 years ago

3 0

Yeah, Tim would have to be 75.

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5a-z=8 -2a+5z =-6 What is the answer

timurjin [86]

Answer:

122 = 7(z + 8) + -5(6 + z) >z = 48

Step-by-step explanation:

122 = 7(8 + z) + -5(6 + z)

122 = (8 * 7 + z * 7) + -5(6 + z)

122 = (56 + 7z) + -5(6 + z)

122 = 56 + 7z + (6 * -5 + z * -5)

122 = 56 + 7z + (-30 + -5z)

Reorder the terms:

122 = 56 + -30 + 7z + -5z

Combine like terms: 56 + -30 = 26

122 = 26 + 7z + -5z

Combine like terms: 7z + -5z = 2z

122 = 26 + 2z

Solving

122 = 26 + 2z

Solving for variable 'z'.

Move all terms containing z to the left, all other terms to the right.

Add '-2z' to each side of the equation.

122 + -2z = 26 + 2z + -2z

Combine like terms: 2z + -2z = 0

122 + -2z = 26 + 0

122 + -2z = 26

Add '-122' to each side of the equation.

122 + -122 + -2z = 26 + -122

Combine like terms: 122 + -122 = 0

0 + -2z = 26 + -122

-2z = 26 + -122

Combine like terms: 26 + -122 = -96

-2z = -96

8 0

3 years ago

Joel can run around a 14– mi track in 81 sec, and Jason can run around the track in 69 sec. If the runners start at the same poi

Alenkasestr [34]

Answer:37.2 sec

Step-by-step explanation:

Given

Joel takes 81 sec for 14 mile long track

speed of joel $\frac{14}{1.35}$ mi/min=10.37 mi/min

Jason takes 69 sec for 14 mi track

so jason speed is $\frac{14}{1.15} mi/min=12.17 mi/min$

if both starts from same spot in opposite direction then

$\frac{x}{10.37}=\frac{14-x}{12.17}$

where x is distance covered by Joel

then x=6.44 miles

therefore time required

$t=\frac{6.44}{10.37}=0.621 min\approx 37.2 sec$

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Y = mx + b

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Answer:

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