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strojnjashka [21]
2 years ago
10

4/7=×/6 what does × =

Mathematics
2 answers:
mart [117]2 years ago
6 0

Answer:

3 3/7 or 24/7

As a decimal: 3.42857143

Step-by-step explanation:

Rewrite the proportion:

4/7 = x/6

Multiply both sides by 6 to isolate x:

(4/7) * 6 = x

Multiply 4 and 6, 7 is multiplied by 1 since 6 = 6/1:

x = 24/7

Rewrite as mixed number:

7 * 3 = 21

24 - 21 = 3

x = 3 3/7

Brainliest, please :) (Hoping to become a genius)

kkurt [141]2 years ago
3 0

Answer:

3.42857143

Step-by-step explanation:

You multiply 4/7 by 6 to isolate the variable

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Convert the equation to slope- intercept and standard form.<br> y-4=5 + 1/2x
Vitek1552 [10]

Answer:

note :

The equation of a linear function in point-slope form is y – y1 = m(x – x1)

The point is A (x1 , y1)


Step-by-step explanation:

in this exercice :  y - 4 = 5 +(1/2)x

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What is the simplified form of the expression k^3(K^7/5)^-5
Vladimir79 [104]
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3 years ago
After refueling, he starts from rest and leaves the pit area with an acceleration whose magnitude is 5.1 m/s2; after 3.6 s he en
Naddik [55]

Answer:

The entering car is going to catch up with the other car after 20.76 seconds.

Step-by-step explanation:

The car leaving the pit area is moving with a constant accelaration, then we can calculate the speed it has when entering the main speedway using the following equation:

s_1=s_0+a*t_1

Where s_0 is the initial speed (which is zero given that the car starts froms rest), a is the car's accelaration and t_1 is the time passed until it reaches the main speedway.

s_1=s_0+a*t_1=a*t_1

s_1=a*t_1=(5.1 \frac{m}{s^2})(3.6s)=18.36 \frac{m}{s}

For an object travelling at a constant accelation, the displacement x at a time t can be calculate using the following equation:

x=x_1+(s_1*t)+\frac{1}{2}at^2 <em>(equation 1)</em>

Let's consider x_1 the point where the first car enters the main speedway. If x_2 is the point where this car catch up with other one after t_2 seconds, we may rewrite the <em>equation 1</em> like this:

x_2=x_1+(s_1*t_2)+\frac{1}{2}a(t_2)^2

x_2=(s_1*t_2)+\frac{1}{2}a(t_2)^2 <em>(equation 2)</em>

In other hand, the second car is travelling at a constant speed (v) when it meets the entering car. Then, its displacement (d_2) after t_2 seconds can be calculated using the following formula:

d_2=v*t_2 <em>(equation 3)</em>

The entering car is going to catch up with the other one when x_2=d_2, so when can find how much time this will require equaling <em>equations 2 and 3</em> and isolating t_2

v*t_2=(s_1*t_2)+\frac{1}{2}a(t_2)^2

0=(s_1*t_2)+\frac{1}{2}a(t_2)^2-v*t_2

[\frac{1}{2}(a*t_2)-(v-s_1)]*t_2=0

\left \{ {{t_2=0} \atop {\frac{1}{2}(a*t_2)-(v-s_1)=0}} \right.

\frac{1}{2}(a*t_2)-(v-s_1)=0

t_2=\frac{2}{a}(v-s_1)=\frac{2}{5.1 \frac{m}{s^2}}(71.3\frac{m}{s}-18.36\frac{m}{s})=20.76 s

So, the entering car is going to catch up with the other car after 20.76 seconds.

8 0
3 years ago
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