What function represents the sequence: 20, 26, 32, 38

How do you use numerical expressions to solve real world problems

Sonja [21]

Ask your teacher .... no kidding but really just convert the numbers

4 0

3 years ago

= 1/2 (u+v)t

RoseWind [281]

The distance travelled by an object is 50 m.

<h3>What is Distance?</h3>

The distance between two points is the length of the path connecting them.

Here, given equation of distance

S = 1/2 (u + v)t

Where s is the distance traveled by an object (in meters)

initial velocity u (in m/s)

final velocity v (in m/s).

t in seconds.

Now, given values are;

u = 8 m/s ; v = 12 m/s ; t = 5 sec.

S = 1/2 (8 + 12)5

S = 1/2 (20)5

S = 100/2

S = 50 m.

Thus, the distance travelled by an object is 50 m.

Learn more about Distance from:

brainly.com/question/15172156

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5 0

2 years ago

Solve each equation by factoring,<br>x² - 2x - 15 = 0

valina [46]

Answer:

x=−3 or x=5

Step-by-step explanation:

x2−2x−15=0

Step 1: Factor left side of equation.

(x+3)(x−5)=0

Step 2: Set factors equal to 0.

x+3=0 or x−5=0

x=−3 or x=5

7 0

3 years ago

Find derivative problem<br> Find B’(6)

dalvyx [7]

Answer:

$B^\prime(6) \approx -28.17$

Step-by-step explanation:

We have:

$\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)$

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

$\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]$

We can move the constant outside:

$\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]$

Now, we will utilize the product rule. The product rule is:

$(uv)^\prime=u^\prime v+u v^\prime$

We will let:

$\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t$

Then:

$\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1$

(The derivative of u was determined using the chain rule.)

Then it follows that:

$\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}$

Therefore:

$\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]$

By simplification:

$\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17$

So, the slope of the tangent line to the point (6, B(6)) is -28.17.

5 0

3 years ago

Where will her cut be located? Round to the nearest tenth. x = (StartFraction m Over m + n EndFraction) (x 2 minus x 1) + x 1 A

Arte-miy333 [17]

Answer:

The correct answer is 25.2 in.

Step-by-step explanation:

It is given that number line goes from 0 to 60 which can be used to represent a ribbon of length = 60 inches.

2 inches of the ribbon are frayed so actual length = 58 inches

Please refer to the attached image for the ribbon.

A is at 0

C is at 60

B is at 2

P is the point to divide the remaining ribbon in the ratio 2:3.

Part AB of the ribbon is frayed.

BP: PC = 2:3

Let BP = 2 $x$ and PC = 3 $x$

Now, BP + PC = BC = 58 = 2 $x$ + 3 $x$ = 5 $x$

So,

$5x =58\\\Rightarrow x =11.6$

BP = $2\times x = 2 \times 11.6 = 23.2\ inches$

Location of the Cut = 2 + 23.2 = <em>25.2 inches</em>

Alternatively, we can use the formula directly:

$x = \dfrac{m} {m + n } (x_2 - x_1) + x_1$

$x_1 = 2\\x_2 = 60$

m: n is the ratio 2:3

$x = \dfrac{2} {2 +3 } (60- 2) + 2\\\Rightarrow 0.4 (58 )+2\\\Rightarrow 23.2+2 \\\Rightarrow \bold{25.2\ inches}$

5 0

3 years ago

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