All categories

3 years ago

Kelly shows all the ways to make 49 as tens and ones. What ways does she show?

Mathematics

2 answers:

Tcecarenko [31]3 years ago

7 0

4 tens and 9 ones 3 tens and 19 ones 2 tens and 29 ones 1 ten and 39 ones HOPE THIS HELPS

ahrayia [7]3 years ago

5 0

There are a bunch of ways to do it but a couple would be 4 tens and 9 ones, or 49 ones. I hope this was helpful.

You might be interested in

Se uma pessoa comprar um aparelho eletrônico em 5 prestações mensais de 304,00 quanto pagará por esse aparelho ?

kobusy [5.1K]

Answer:

1520

Step-by-step explanation:

Pela pergunta acima, sabemos que:

1 parcela mensal = 304

5 parcelas mensais = x

Multiplicação cruzada

x = 5 prestações mensais × 304/1 prestações mensais

= 1520

Portanto, o valor que ele pagou pelo dispositivo é 1520

6 0

3 years ago

What is the slope of a graph?

e-lub [12.9K]

Answer:

Image result for what is the slope of a graph?

The slope of a line is rise over run. Learn how to calculate the slope of the line in a graph by finding the change in y and the change in x.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

What is the value of p?

maks197457 [2]

Can you show all the answer choices, so I know if I am correct or not?

7 0

3 years ago

Help with a,b,c answers

Ket [755]

The answers For A,B,C.
A.2
B.2.25
C.1

6 0

3 years ago

Y=arccos(1/x)<br><br> Please help me do them all! I don’t know derivatives :(

Stells [14]

Answer:

$f(x) = {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x \rightarrow \: x = sec \: y\\ \frac{dx}{dx} = \frac{d(sec \: y)}{dx} \\ 1 = \frac{d(sec \: y)}{dx} \times \frac{dy}{dy} \\ 1 = \frac{d(sec \: y)}{dy} \times \frac{dy}{dx} \\1 = tan \: y.sec \: y. \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{1}{tan \: y.sec \: y} \\ \frac{dy}{dx} = \frac{1}{ \sqrt{( {sec}^{2} \: y - 1}) .sec \: y} \\ \frac{dy}{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx} = \frac{1}{ |5x | \sqrt{25 {x }^{2} - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} }$

4 0

2 years ago