All categories

3 years ago

PLZZZZ HLPPPPP MEEEEEEEE

Mathematics

1 answer:

kati45 [8]3 years ago

4 0

Answer: y ≥ (3/5)*x - 3

Step-by-step explanation:

In the graph, we can see that we are above a bold line, that goes through the points (0, -3) and (5, 0)

First, let's find the equation for this line:

y = a*x + b

the value of a is the slope and is equal to:

a = (0 - (-3))/(5 -0) = 3/5

and the value of b is the point where the line intersects the y-axis, in this case, b = -3

then our line is

y = (3/5)*x - 3

As the shaded part is above the line, this equality represents the minimum value that y can take for a given x, and because the line is not a doted line, we know that the equality is valid, so we must use the ≥ symbol.

y ≥ (3/5)*x - 3

You might be interested in

Decide whether the rates are equivalent 30 beats per 20 seconds, 90 beats per 60 seconds. Show your work.

siniylev [52]

Ghjyhuytrdruiouyeroiufduiidfghopoiuytddfghjk

8 0

3 years ago

Read 2 more answers

Round 249798.828806 to the nearest ten.

Artyom0805 [142]

249798.8

The nearest ten

5 0

2 years ago

Read 2 more answers

All of the following are rational numbers: .5, 7, V16<br> O True<br> O False

wel

Answer:

Your answer is: A) True

( I say true because 1) .5 2) 7 are rational) ( I dont know about v16, if that is what it is supposed to say, then I cant quite tell)

A rational number is any integer, fraction, terminating decimal, or repeating decimal.

Step-by-step explanation:

Hope this helped : )

4 0

2 years ago

Helllp,??????????????

makkiz [27]

If the first question mark is 3 then the next is 13 than the next still 13 and the right one 26 and then lastly x is equally to 2
Or
3x+9+10x=33
13x+9=33
13x=26
X=2

3 0

3 years ago

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Otrada [13]

I guess the "5" is supposed to represent the integral sign?

$I=\displaystyle\int_1^4\ln t\,\mathrm dt$

With $n=10$ subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

$\ell_i=1+\dfrac{3(i-1)}{10}$

and right endpoints are given by

$r_i=1+\dfrac{3i}{10}$

where $1\le i\le10$ .

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, $\dfrac{4-1}{10}=\dfrac3{10}$ , and "bases" equal to the values of $\ln t$ at both endpoints of each subinterval. The area of the trapezoid over the $i$ -th subinterval is

$\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)$

Then the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}$

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of $\ln t$ at the average of the subinterval's endpoints, $\dfrac{\ell_i+r_i}2$ . The area of the rectangle over the $i$ -th subinterval is then

$\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}$

so the integral is approximately

$I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}$

c. For Simpson's rule, we find a quadratic interpolation of $\ln t$ over each subinterval given by

$P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}$