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2 years ago

PLEASE HELP!!!

Mathematics

1 answer:

Paladinen [302]2 years ago

4 0

Answer:

d (w) = 0.1w² – w+ 5

Step-by-step explanation:

d(w) = c(w) - a(w)

= -0.3w²+2w+13 -(-0.4w²+3w+8)

= -0.3w²+0.4w²+2w-3w+13-8

= 0.1w²-w+5

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Explain how bacteria are both helpful and harmful to out bodies

Pavel [41]

Answer:

Step-by-step explanation:

Bacteria is helpful because it helps protect the cells in your intestines. Its, harmful because itI can cause a disease in your lungs.

Hope this helps!

8 0

2 years ago

Use the graph below to determine a1 and d for the sequence. graphed sequence showing point 1, negative 10, point 2, negative 7,

Sonja [21]

Answer:

$a_1=-10$

$d=3$

Step-by-step explanation:

we know that

In an <u><em>Arithmetic Sequence</em></u> the difference between one term and the next is a constant, and this constant is called the common difference (d)

In this problem we have the ordered pairs

$(1,-10),(2,-7),(3,-4),(4,-1),(5,2),(6,5)$

Let

$a_1=-10\\a_2=-7\\a_3=-4\\a_4=-1\\a_5=2\\a_6=5$

Find the difference between one term and the next

$a_2-a_1=-7-(-10)=3$

$a_3-a_2=-4-(-7)=3$

$a_4-a_3=-1-(-4)=3$

$a_5-a_4=2-(-1)=3$

$a_6-a_5=5-2=3$

The difference between one term and the next is a constant

This constant is the common difference

The sequence graphed is an Arithmetic Sequence

therefore

The first term is $a_1=-10$

The common difference is equal to $d=3$

4 0

3 years ago

Please please help me

Alex73 [517]

Answer:

x = 5.5

Step-by-step explanation:

Given 2 secants intersecting the circle from a point outside the circle then

The product of the external part and the entire part of one secant is equal to the product of the external part and the entire part of the other secant, that is

x(x + 14) = 6(6 + 12)

x² + 14x = 6 × 18 = 108 ( subtract 108 from both sides )

x² + 14x - 108 = 0 ← in standard form

with a = 1, b = 14 and c = - 108

Using the quadratic formula to solve for x

x = ( - b ± $\sqrt{b^2-4ac}$ ) / 2a

= ( - 14 ± $\sqrt{14^2-(4(1)(-108)}$ ) / 2

= ( - 14 ± $\sqrt{196+432}$ ) / 2

= ( - 14 ± $\sqrt{628}$ ) / 2

x = $\frac{-14-\sqrt{628} }{2}$ or x = $\frac{-14+\sqrt{628} }{2}$

x = - 19.5 or x = 5.5 ( to 1 dec. place )

However x > 0 ⇒ x = 5.5

6 0

2 years ago

NEED ANSWER ASAP!!! Do the data in this table show a function? If you switch the input and the output values is it a function?

ser-zykov [4K]

Answer:

Step-by-step explanation:

no input can have diffrent outputs if you change the input values then yes it is a function hope this helps god bless

4 0

2 years ago

Does anyone know how to graph trigonometric functions for algebra 2?

anastassius [24]

Solution:

Give the following below

$\begin{gathered} amplitude=3 \\ period=\frac{\pi}{2} \\ midline,\text{ }y=-1 \\ Passing\text{ through the point }(0,2) \end{gathered}$

To find the cosine function, we will apply the general formula for cosine function below

$\begin{gathered} f(x)=a\cos(bx-c)+d \\ Where \\ a\text{ is the amplitude} \\ b\text{ represents the speed of the cycle} \\ d\text{ is the vertical shift} \\ \frac{c}{b}\text{ is the phase shift \lparen horizontal shift\rparen} \\ Period=\frac{2\pi}{b} \end{gathered}$

To find the value of b

$\begin{gathered} Period=\frac{\pi}{2} \\ \frac{2\pi}{b}=\frac{\pi}{2} \\ Crossmultiply \\ b\times\pi=2\times2\pi \\ b=\frac{4\pi}{\pi} \\ b=4 \end{gathered}$ $\begin{gathered} a=3 \\ b=4 \\ d=-1 \\ c=0 \end{gathered}$

Substitute the values of the variables into the general formula for cosine function

$\begin{gathered} f(x)=a\cos(bx-c)+d \\ f(x)=3\cos(4x-0)-1 \\ f(x)=3\cos4x-1 \end{gathered}$

Applying a graphing tool,

The graph of one cycle of the function is shown below

4 0

10 months ago