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3 years ago

Which equation is a true proportion please help

Mathematics

1 answer:

NNADVOKAT [17]3 years ago

3 0

Answer: The 3rd equation is a true proportion

Brainliest please! <3

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The center of a circle is at the origin on a coordinate grid. A line with a positive slope intersects the circle at (0, 7). Whic

denis23 [38]

The true statement is the circle has the radius equal to 7.

<h3>What is Equation of circle?</h3>

The general equation for a circle is (x-h)² + (y-k)² = r², where ( h, k ) is the center and r is the radius.

As, standard form of equation of circle is

(x-h)² + (y-k)² = r²

At origin, equation of circle will be

(0-h)² + (0-k)² = r²

r² = x² + y²

Now, The line intersects the circle at (0,7)

r² = 0² + 7²

r² =49

r=7.

Hence, the circle has the radius equal to 7.

Learn more about this concept here:

brainly.com/question/11702700

#SPJ1

3 0

2 years ago

A JET TRAVELS 460 MILES IN 2 HOURS?

Tanzania [10]

Find how fast it travels in one hour:

460 miles / 2 hours = 230 miles per hour.

A) Multiply the speed per hour by the number of hours:

230 x 8 = 1840 miles in 8 hours.

B) 230 x 20 = 4,600 miles in 20 hours.

C) 460 miles / 2 hours = 230 miles per hour.

8 0

3 years ago

Read 2 more answers

Pls tell/show me how u did it<br><br>Also the answer that there is worng

goblinko [34]

Answer:

y=4/3x+2

Step-by-step explanation:

y=mx+b

b is y intercept, and in this case, 2

m is the slope, and in this case, 4/3

Hope this answers your question.

3 0

2 years ago

Read 2 more answers

Five members of the soccer team and five members of the track team ran the 100-meter dash. Their times are listed in the table b

viva [34]

The last one....0.74

5 0

3 years ago

Read 2 more answers

Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 i

rosijanka [135]

It looks like the ODE is

$y'+5y=\begin{cases}0&\text{for }0\le t$

with the initial condition of $y(0)=4$ .

Rewrite the right side in terms of the unit step function,

$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t$

In this case, we have

$\begin{cases}0&\text{for }0\le t$

The Laplace transform of the step function is easy to compute:

$\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s$

So, taking the Laplace transform of both sides of the ODE, we get

$sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s$

Solve for $Y(s)$ :

$(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}$

We can split the first term into partial fractions:

$\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs$

If $s=0$ , then $1=5a\implies a=\frac15$ .

If $s=-5$ , then $1=-5b\implies b=-\frac15$ .

$\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}$

$\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}$

Take the inverse transform of both sides, recalling that

$Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)$

where $F(s)$ is the Laplace transform of the function $f(t)$ . We have

$F(s)=\dfrac1s\implies f(t)=1$

$F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}$

We then end up with

$y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}$

3 0

4 years ago