Ask question

Ask question

All categories

2 years ago

15

Solve for x 2/3=x/4

0" id="TexFormula1" title=" \frac{2}{3 } = \frac{x}{4 } " alt=" \frac{2}{3 } = \frac{x}{4 } " align="absmiddle" class="latex-formula">

1 answer:

balandron [24]2 years ago

3 0

To solve this, you cross multiply, then divide. In this case, we multiply 4 and 2, then divide by 3.

4 x 2 = 8
8 / 3 = 2.66

So, x is 2.66

You might be interested in

What is the range of the relation? {(1, 2), (2, 4), (3, 2), (4, 6)} A.{2, 4} B.{1, 2, 3, 4, 6} C.{2, 4, 6} D.{1, 2, 3, 4}

tamaranim1 [39]

$(1)\ \ \ \ range:\ \ \ \{2, 4, 6\}\ \ \ \Rightarrow\ \ \ Ans.\ C\\\\(2)\ \ \ function:\ \ \ \{(1, 2); (2, 2); (3, 2); (4, 2)\}\ \ \ \Rightarrow\ \ \ Ans.\ D\\\\(3)\ \ \ f(4)=2-3\cdot4=2-12=-10\ \ \ \Rightarrow\ \ \ Ans.\ C\\\\(4)\ \ \ direct:\ \ \ y=ax\ \ \ \Rightarrow\ \ \ y= \frac{1}{3} x\ \ \ \Rightarrow\ \ \ Ans.\ B\\\\(5)\ \ \ f(x)=ax\ \ \ and\ \ \ f(4)=12\\.\ \ \ \ \ \Rightarrow\ \ \ 12=a\cdot4\ \ \ \Rightarrow\ \ \ a=3\ \ \ \Rightarrow\ \ \ f(x)=3x\ \ \ \Rightarrow\ \ \ Ans.\ (?)\\\\$

$(6)\ \ \ f(x)=ax^2\ \ \ and\ \ \ f(2)=48\\.\ \ \ \Rightarrow\ \ 48=a\cdot2^2\ \ \Rightarrow\ \ a=48:4=12\ \ \Rightarrow\ \ f(x)=12x^2\ \Rightarrow\ \ Ans.\ C\\\\(7)\ \ \ inversely:\ \ \ f(x)= \frac{a}{x} \ \ \ and\ \ \ f(2)=4\\.\ \ \ \Rightarrow\ \ \ 4= \frac{a}{2} \ \ \ \Rightarrow\ \ \ a=4\cdot2=8\ \ \ \Rightarrow\ \ \ y= \frac{8}{x}\ \ \ \Rightarrow\ \ \ Ans.\ B$

5 0

3 years ago

What are the x-intercept and y-intercept of the graph of y=13x−6 ?

mestny [16]

Answer:

x intercept is 6/13

y intercept is -6

Step-by-step explanation:

x intercepted when y is zero.

y intecepted when x is zero...

3 0

3 years ago

If the simple interest on $6,000 for 3 years is $1,620, then what is the interest rate?

Volgvan

Answer:

Interest rate: 9%

Step-by-step explanation:

Using the simple interest formula: A=P(1+rt)

A=Final amount

P=Principal amount

r=rate

t=time (years)

We just need to put in the information we have from the question and solve for r.

7,620=6000(1+r3)

7,620/6000=1+r3

1.27=1+r3

0.27=r3

r=0.09

r=9%

3 0

2 years ago

A particle moves on a straight line and has acceleration a(t)=24t+2. Its position at time t=0 is s(0)=3 and its velocity at time

user100 [1]

Answer:

It's position at time t = 5 is 593.

Step-by-step explanation:

The velocity v(t) is the integral of the acceleration a(t)

The position s(t) is the integral of the velocity v(t)

We have that:

The acceleration is:

$a(t) = 24t + 2$

Velocity:

$v(t) = \int {a(t)} \, dt = \int {24t + 2} \, dt = 12t^{2} + 2t + K$

K is the initial velocity, that is v(0). Since V(0) = 13, K = 13

Then

$v(t) = 12t^{2} + 2t + 13$

Position:

$s(t) = \int {s(t)} \, dt = \int {12t^{2} + 2t + 13} \, dt = 4t^{3} + t^{2} + 13t + K$

Since s(0) = 3

$s(t) = 4t^{3} + t^{2} + 13t + 3$

What is its position at time t=5?

This is s(5).

$s(t) = 4t^{3} + t^{2} + 13t + 3$

$s(5) = 4*5^{3} + 5^{2} + 13*5 + 3$

$s(5) = 593$

It's position at time t = 5 is 593.

3 0

2 years ago

Alinara [238K]

It is the first answer, loss of $701.80

5 0

2 years ago