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2 years ago

(Practice test) algebra 2 What is the period of the function. Please help I don’t understand this

Mathematics

1 answer:

loris [4]2 years ago

4 0

Answer:

2π

Step-by-step explanation:

period is the distance between maximum to maximum or minimum to minimum.

π/3 - - 5π/3

= 2π

here the period is 2π

// appreciate it if u mark this as brainliest and change the questioning heading or u gonna get reported lols//

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I need help (please show your work)

oksano4ka [1.4K]

Answer:

the solution is (-4, 3)

Step-by-step explanation:

if you were to graph y = -7/4x - 4 you could plot your first point at (0, -4) and get your second point by going up 7 units (from (0, -4) and to the left by 4

if you were to graph y = -1/4x + 2 you could plot your first point at (0, 2) and get your second point by going up 1 unit (from (0, 2) and to the left by 4

the second point plotted for each equation will be the point of intersection, (-4, 3)

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3 years ago

The ratio of two numbers is 1:4, and the sum of these numbers is 40. Find the numbers

Irina-Kira [14]

X + 4x = 40

5x = 40

5x/5 = 40/5

x = 8

1x 8 = 8

4x 8 = 32

Check=

8 + 32 = 40

The numbers are 8 and 32

3 0

2 years ago

1 - 2/4=?????<br>I forgot how too do these T-T

Tema [17]

Answer:

1/2

Step-by-step explanation:

you can simply the fraction making it 1/2 and 1 - 1/2 = 1/2

3 0

2 years ago

Read 2 more answers

Mark Brainliest what is 2/3 + 7/8 Please help

Pachacha [2.7K]

1 13/24 and that's it

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3 years ago

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On the 1st January 2014 Carol invested some money in a bank account.

Ghella [55]

Answer:

$\large \boxed{\text{\pounds 23 360.00}}$

Step-by-step explanation:

The formula for the accrued amount from compound interest is

$A = P \left(1 + \dfrac{r}{n}\right)^{nt}$

1. Amount in account on 1 Jan 2015

(a) Data:

a = £23 517.60

r = 2.5 %

n = 1

t = 1 yr

(b) Calculations:

r = 0.025

$\begin{array}{rcl}23517.60 & = & P\left (1 + \dfrac{r}{n}\right)^{nt}\\\\& = & P\left (1 + \dfrac{0.025}{1}\right)^{1\times1}\\\\& = & P (1 + 0.025)\\ & = & 1.025 P\\P & = & \dfrac{23517.60 }{1.025} \\\\& = & 22 944.00 \\\end{array}$

The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.

She must have had £23 944 in her account on 1 Jan 2015.

(2) Amount originally invested

(a) Data

A = £23 944.00

$\begin{array}{rcl}23 944.00 & = & 1.025 P\\P & = & \dfrac{23 944.00 }{1.025} \\\\& = & \mathbf{23 360.00} \\\end{array}\\\text{Carol originally invested $\large \boxed{\textbf{\pounds23 360.00}}$ in her account.}$

3. Summary

1 Jan 2014 P = £23 360.00

1 Jan 2015 A = 23 944.00

Withdrawal = <u> -1 000.00 </u>

P = 22 944.00

1 Jan 2016 A = £23 517.60

5 0

2 years ago