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2 years ago

Uhhhhhh????????????!

Mathematics

1 answer:

mariarad [96]2 years ago

8 0

Answer:

2y + 3 = 4y + 2; y = 1/2

Step-by-step explanation:

to find this equation we can use the ys and the 1s to create an equation

in the left side

there are 2 ys and 3 1s

your equation is

2y + 3

on the other side

it is

4 ys and 2 1s

4y + 2 is your equation

then yo uset them equal to each other

2y + 3 = 4y + 2

if you must solve

subtract 2y and 2 from both sides

1 = 2y

then divide both sides by 2

y = 1/2 is your answer

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How can you write -1/4x-9 on a graph

Dmitrij [34]

Answer:

Slope : -1/4

y-intercept : -9

x... 0 , 1

y... -9 , -37/4

Step-by-step explanation: Graph the line using the slope and y-intercept, or two points. GRAPH IS DOWN BELOW!

Hope this helps you out.

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3 years ago

Find the missing angle. PLEASE HELP!! I WILL GIVE 100 POINTS!!

mariarad [96]

Answer:

50°

Step-by-step explanation:

90°-40°=50°

The square on the angle means 90°.

Hope this helps! :D

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2 years ago

The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+10, where t is measured in s

oksian1 [2.3K]

Answer:

The maximum acceleration over that interval is $A(6) = 28$ .

Step-by-step explanation:

The acceleration of this car is modelled as a function of the variable $t$ .

Notice that the interval of interest $0 \le t \le 6$ is closed on both ends. In other words, this interval includes both endpoints: $t = 0$ and $t= 6$ . Over this interval, the value of $A(t)$ might be maximized when $t$ is at the following:

One of the two endpoints of this interval, where $t = 0$ or $t = 6$ .
A local maximum of $A(t)$ , where $A^\prime(t) = 0$ (first derivative of $A(t)\!$ is zero) and $A^{\prime\prime}(t)$ (second derivative of $\! A(t)$ is smaller than zero.)

Start by calculating the value of $A(t)$ at the two endpoints:

$A(0) = 10$ .
$A(6) = 28$ .

Apply the power rule to find the first and second derivatives of $A(t)$ :

$\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}$ .

$\displaystyle A^{\prime\prime}(t) = 6\, t - 15$ .

Notice that both $t = 1$ and $t = 4$ are first derivatives of $A^{\prime}(t)$ over the interval $0 \le t \le 6$ .

However, among these two zeros, only $t = 1\!$ ensures that the second derivative $A^{\prime\prime}(t)$ is smaller than zero (that is: $A^{\prime\prime}(1) < 0$ .) If the second derivative $A^{\prime\prime}(t)\!$ is non-negative, that zero of $A^{\prime}(t)$ would either be an inflection point (if $A^{\prime\prime}(t) = 0$ ) or a local minimum (if $A^{\prime\prime}(t) > 0$ .)

Therefore $\! t = 1$ would be the only local maximum over the interval $0 \le t \le 6\!$ .

Calculate the value of $A(t)$ at this local maximum:

$A(1) = 15.5$ .

Compare these three possible maximum values of $A(t)$ over the interval $0 \le t \le 6$ . Apparently, $t = 6$ would maximize the value of $A(t)\!$ . That is: $A(6) = 28$ gives the maximum value of $\! A(t)$ over the interval $0 \le t \le 6\!$ .

However, note that the maximum over this interval exists because $t = 6\!$ is indeed part of the $0 \le t \le 6$ interval. For example, the same $A(t)$ would have no maximum over the interval $0 \le t < 6$ (which does not include $t = 6$ .)