Find the sum. 6(2k – 2) + (7k + 5)

Mathematics

2 answers:

zubka84 [21]4 years ago

6 0

Answer:

Your answer is: 19k - 7

Simplify the expression.

Step-by-step explanation:

Hope this helped : )

Tatiana [17]4 years ago

5 0

Answer is 19k-7

6(2k-2) + (7k+5)
12k-12+7k+5
19kx12+5
19k-7

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Two cars start out together from the same place. they travel in opposite directions, with one of them traveling 5 miles per hou

Mashutka [201]

<u>Annotation</u>
General formula for distance-time-velocity relationship is as following
d = v × t
The velocity of the first car will be v₁, the time is 2 hours, the distance will be d₁.
The velocity of the second car will be v₂, the time is 2 hours, the distance will be d₂.

One of them traveling 5 miles per hour faster than the others. That means the velocity of the first car is 5 miles per hour more than the velocity of the second car.
v₁ = v₂ + 5 (first equation)

The distance of the two cars after two hours will be 262 miles apart. Because they go to opposite direction, we could write it as below.
d₁ + d₂ = 262 (second equation)

Plug the d-v-t relationship to the second equation
d₁ + d₂ = 262
v₁ × t + v₂ × t = 262
v₁ × 2 + v₂ × 2 = 262
2v₁ + 2v₂ = 262

Plug the v₁ as (v₂+5) from the first equation
2v₁ + 2v₂ = 262
2(v₂ + 5) + 2v₂ = 262
2v₂ + 10 + 2v₂ = 262
4v₂ + 10 = 262
4v₂ = 252
v₂ = 252/4
v₂ = 63
The second car is 63 mph fast.

Find the velocity of the first car, use the first equation
v₁ = v₂ + 5
v₁ = 63 + 5
v₁ = 68
The first car is 68 mph fast.

Answer
$\boxed{\boxed{ v_{1}=68mph} }$
$\boxed{\boxed{ v_{2}=63mph} }$

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Tyler and Pearl went on a 20km bike ride that lasted for 3 hours. Because of steep hills or the trail, they walked for most of t

noname [10]

Answer: 1 hour

Step-by-step explanation:

Let the distance they walked be $d_1$ in time $t_1$ and the distance they walked be $d_2$ in time $t_2$ ,then

Walking speed= $\frac{d_1}{t_1}=4$

$\\\Rightarrow\ d_1=4t_1$

Riding speed= $\frac{d_2}{t_2}=12$

$\\\Rightarrow\ d_2=12t_1$

Also the total time taken=3 hours

∴ $t_1+t_2=3\\\Rightarrow\ t_2=3-t_1$ ...(1)

Since the total distance= 20 km

∴ $d_1+d_2=20\\\Rightarrow4t_1+12t_2=20\\\Rightarrow4t_1+12(3-t_1)=20..............[\text{from (1)}]\\\Rightarrow4t_1+36-12t_1=20\\\Rightarrow-8t_1=-8\\\Rightarrow\ t_1=1$