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3 years ago

Solve for x 3x+4=9x+3

Mathematics

2 answers:

viva [34]3 years ago

8 0

Answer:

x=0.17

Step-by-step explanation:

SpyIntel [72]3 years ago

3 0

1. subtract 3x from both sides

2. subtract 3 from both sides

3. divide both sides by 6

x= 1 /6

Use math papa its a algebra calculator

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John's dog eats 8 ounces of dog food each day. John bought 20-pound bag of dog food. How many 8 - ounces are in 28- pounds of do

aleksley [76]

8x3= 24...
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hope this helps!! HAGD!

5 0

3 years ago

Read 2 more answers

Will mark brainliest if correct :)

Finger [1]

Answer:

d = 12

c = 4

so I think d > c

Step-by-step explanation:

1 x 12 = 12

0.75 x 4 = 3

12 + 3 = 15

8 0

3 years ago

Graph this rational equation. Identify the points of discontinuity, holes, vertical asymptotes, x-intercepts, and horizontal asy

irakobra [83]

Step-by-step explanation:

We have given,

A rational function : f(x) = $\frac{x-2}{x-4}$

W need to find :

Point of discontinuity : - At x = 4, f(x) tends to reach infinity, So we get discontinuity point at x =4.

For no values of x, we get indetermined form (i.e $\frac{0}{0}$ ), Hence there is no holes

Vertical Asymptotes:

Plug y=f(x) = ∞ in f(x) to get vertical asymptote {We can us writing ∞ = $\frac{1}{0}$ }

i.e ∞ = $\frac{x-2}{x-4}$

or $\frac{1}{0}=\frac{x-2}{x-4}$

or x-4 =0

or x=4, Hence at x = 4, f(x) has a vertical asymptote

X -intercept :

Plug f(x)=0 , to get x intercept.

i.e 0 = $\frac{x-2}{x-4}$

or x - 2 =0

or x = 2

Hence at x=2, f(x) has an x intercept

Horizontal asymptote:

Plug x = ∞ in f(x) to get horizontal asymptote.

i.e f(x) = $\frac{x-2}{x-4}$ = $\frac{x(1-\frac{2}{x} )}{x(1-\frac{4}{x} )}$

or f(x) = $\frac{(1-\frac{2}{∞} )}{(1-\frac{4}{∞} )}$

or f(x) = 1 = y

hence at y =f(x) = 1, we get horizontal asymptote

4 0

3 years ago

(-9, -5) and (-1, -9)<br> find the midpoint

irga5000 [103]

Answer:

(-5,-7)

Step-by-step explanation:

6 0

2 years ago

Which of the following are identities? Check all that apply. A. cot2x + 1 = csc2x B. tan2x = 1 - sec2x C. sin2x = 1 + cos2x D. s

kirill115 [55]

I'll assume those are squares. We know D is an identity:

$\sin^2 x + \cos ^2 x = 1$

Dividing through by $\sin ^2 x$

$1 + \cot^2 x = \csc^2 x$

That's A.

Dividing the original through by $\cos ^2 x$

$\tan^2 x + 1 = \sec^2 x$

Not quite B, wrong sign on tangent.

C has the wrong sign on cosine squared as well.

Identities: A & D

4 0

3 years ago