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3 years ago

I NEED HELPPPP

Mathematics

1 answer:

Anna007 [38]3 years ago

7 0

I’m sorry this is all I can say! So she has 28 points. But what do we really have to record?

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The LCM and the GCF of two numbers is 120 and 8 respectively. If the first number is 24, find the other number

Ivanshal [37]

Answer: The other no. is 5

Step-by-step explanation: Let the other no. be x

x = LCM/First no.

x = 120/24 = 5

The other no. is 5

7 0

3 years ago

Which function is graphed below?

hoa [83]

Option B. y = 3 (one-third) Superscript x is the function that is graphed in this question.

<h3>How to solve the problem</h3>

We have exponential functions to be of the form abˣ

This can be written in the form of f(x) = $3\frac{1}{3} ^x$

So it crosses through the point (0, 3).

From the way that the curve passes through the circle, we can see that the it is said to pass through at (0, 3) and approaches y = 0 in quadrant 1

Hence this would put our answer to be y = 3 (one-third) Superscript x

Read more on coordinate planes here

brainly.com/question/27975730?referrer=searchResults

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4 0

2 years ago

Read 2 more answers

Given the net, what is the surface area of the rectangular prism below?

Vera_Pavlovna [14]

Answer: 56

Step-by-step explanation: I got this answer by taking each rectangles legth and width and multiplying those, then I added all of the final answers together to get 56

8 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

HELP ME ITS MATH PLEASE

REY [17]

Answer:

so the first one unselect

the second one check it

the third one unselect

the fourth one check it

the fifth one check it

4 0

3 years ago

Read 2 more answers