Ask question

Ask question

All categories

andrezito [222]

3 years ago

12

Drag each label to the correct location on the graph. Not all labels will be used.

1 answer:

mamaluj [8]3 years ago

4 0

Answer:

Not all labels will be used. ... They charge a certain rate for the rental of a sailboat and an additional ... according to the length of time for which the tourists rent the boat.

You might be interested in

If you don’t know then don’t say anything.

tino4ka555 [31]

Answer:

in total there are 80 subjects. I don't know what class width is though.

Step-by-step explanation:

Look at the graph

3 0

3 years ago

How many 3/4 are in 6?

beks73 [17]

Answer: There are 8 3/4 in 6

You can use a rule of three:
If in 3/4 there is 1 (3/4), How many 3/4 are in 6?:

3/4 → 1
6 → x

x=(6)(1)/(3/4)→
x=6/(3/4)→
x=(6/1)/(3/4)→
x=(6/1)(4/3)→
x=(6)(4)/[(1)(3)]→
x=24/3→
x=8

Answer: There are 8 3/4 in 6

3 0

3 years ago

Read 2 more answers

Expanding logarithmic Expression In Exercise,Use the properties of logarithms to rewrite the expression as a sum,difference,or m

wel

Answer:

$\frac{1}{2}ln(2x)-\frac{1}{2}ln(x^2-1)$

Step-by-step explanation:

In 2x/(x^2 - 1)1/2

$ln(\frac{2x}{x^2-1} )^\frac{1}{2}$

Apply the property of natural log

ln x^m = m ln(x) move the exponent before ln

$ln(\frac{2x}{x^2-1} )^\frac{1}{2}$

$\frac{1}{2}ln(\frac{2x}{x^2-1})$

ln(m/n)= ln m - ln n

$\frac{1}{2}(ln(2x)-ln(x^2-1))$

multiply 1/2 inside the terms

$\frac{1}{2}ln(2x)-\frac{1}{2}ln(x^2-1)$

6 0

3 years ago

A circle’s circumference is approximately 44 cm.

geniusboy [140]

Answer:

2fweds

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers

A sequence of numbers follows the rule: multiply the previous number by -2 and add 3. The fourth term in the sequence is -7. Giv

nevsk [136]

Answer:

Next 3 terms: 17, -31, 65

3 terms before -7: 2, -1, 5

Step-by-step explanation:

Fourth term in the sequence = -7

Rule for finding the next term:

$a_n=-2\times a_{n-1}+3$

Therefore, the fifth term:

$a_5=-2\times a_4+3\\\Rightarrow a_5=-2\times -7+3\\\Rightarrow \bold{a_5=17}$

Sixth term:

$a_6=-2\times a_5+3\\\Rightarrow a_6=-2\times 17+3\\\Rightarrow \bold{a_6=-31}$

Seventh term:

$a_7=-2\times a_6+3\\\Rightarrow a_7=-2\times -31+3\\\Rightarrow \bold{a_7=65}$

Let the three terms that came before -7 be: $x, y, z$

As per question statement:

$a_n=-2\times a_{n-1}+3$

$-7 = -2\times z+3\\\Rightarrow -2z=-10\\\Rightarrow z=5$

Again applying the given rule:

$5 = -2\times y+3\\\Rightarrow y =-1$

Again applying the given rule:

$-1 = -2\times x+3\\\Rightarrow x =2$

Therefore the 3 terms before -7 are:

2, -1, 5

5 0

3 years ago