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3 years ago

Using the expressions above, find the cost to the family of each company's phone plan if:

Mathematics

1 answer:

BabaBlast [244]3 years ago

6 0

Answer:

company B

Step-by-step explanation:

the company is good!

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Choose the linear inequality that describes the graph. The gray area represents the shaded region.

jek_recluse [69]

Answer:

2x + 3y ≥ 5

Step-by-step explanation:

See the graph attached.

The bold straight line passes through the points (1,1) and (4,-1).

Therefore, the equation of the straight line will be

$\frac{y + 1}{- 1 - 1} = \frac{x - 4}{4 - 1}$

⇒ 3(y + 1) = - 2(x - 4)

⇒ 3y + 3 = - 2x + 8

⇒ 2x + 3y = 5 ............. (1)

Now, the shaded region i.e. the solution to the inequality does not include the origin(0,0).

So, putting x = 0 and y = 0 in the equation (1) we get, 0 < 5

Therefore, the inequality equation is 2x + 3y ≥ 5 (Answer)

6 0

3 years ago

If you wanted to make the graph of y = 8x - 10 less steep, as well as shift the

azamat

Answer:

4x-15=y

Step-by-step explanation:

The higher the rise over the run the more steep your slope will be thus 4x is less steep than 8x because it does not rise as much.

The y intercept shifted downward means anything (-10>b) can be plausible for b.

3 0

3 years ago

If f(x)=4x-20 what is f(4)

Rudik [331]

Answer:

Step-by-step explanation:

f(x) = 4x - 20

f(4) → put x = 4 to the equation of the function:

f(4) = 4(4) - 20 = 16 - 20 = -4

3 0

3 years ago

Read 2 more answers

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

Triangle BAC was rotated 90° clockwise and dilated at a scale factor of 2 from the origin to create triangle XYZ. Based on these

ololo11 [35]

Answer:b=x

Step-by-step explanation:

8 0

2 years ago