1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Andrews [41]
3 years ago
12

Consider the function g(x) = sqrt(x + 4) - 6 which function has the same y-intercept as this function?

Mathematics
1 answer:
Inga [223]3 years ago
7 0

Answer:

a)

f(x) = \frac{4}{x-8} - 3.5

Step-by-step explanation:

The y intercept of g(x) is the value of g when x = 0.

In this problem

g(x) = \sqrt{x+4} - 6

The y-intercept is

g(0) = \sqrt{0+4} = \sqrt{4} - 6 = 2 - 6 = -4

a)

f(x) = \frac{4}{x-8} - 3.5

The y-intercept is:

f(0) = \frac{4}{0-8} - 3.5 = -0.5 - 3.5 = -4

This is the correct answer

b)

f(x) = -\frac{4}{x} - 1

The y-intercept is

f(0) = -\frac{4}{0} - 1

This is an asymptote

c)

f(x) = 2|x+2|

The y-intercept is

f(0) = 2|0+2| = 4

d)

f(x) = |x - 4|

The y-intercept is

[tex]f(0) = |0-4| = |-4| = 4.

You might be interested in
Identify the figure and denote it by its appropriate symbol
Gekata [30.6K]
B: ray
Rays are a linear formation that start at a point and continue forever in a direction. The photo shows a point on one end, and an arrow on the other indicating that it starts at the point and continues forever past the arrow.
5 0
2 years ago
Flashback! Do you remember what these numbers tell you? The 10 means it will pay only $10,000 maximum for injury to any one pers
iogann1982 [59]

Answer:

  • 10 -- $10k to any one person
  • 20 -- $20k to all persons, total
  • 5 -- $5k for property damage

  all are <em>maximum</em> values

Step-by-step explanation:

Insurance contracts specify <em>maximum</em> payments, not <em>minimum</em> payments. The numbers mean ...

The 10 means it will pay only $10,000 maximum for injury to any one person in an accident.

The 20 means it will pay a maximum of $20,000 total to people injured in one accident.

The 5 means that it will pay a maximum of $5,000 in property damage for one accident.

6 0
3 years ago
Evaluate the limit
wel

We are given with a limit and we need to find it's value so let's start !!!!

{\quad \qquad \blacktriangleright \blacktriangleright \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}

But , before starting , let's recall an identity which is the <em>main key</em> to answer this question

  • {\boxed{\bf{a^{2}-b^{2}=(a+b)(a-b)}}}

Consider The limit ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}

Now as directly putting the limit will lead to <em>indeterminate form 0/0.</em> So , <em>Rationalizing</em> the <em>numerator</em> i.e multiplying both numerator and denominator by the <em>conjugate of numerator </em>

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}\times \dfrac{\sqrt{x}+\sqrt{3\sqrt{x}-2}}{\sqrt{x}+\sqrt{3\sqrt{x}-2}}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-\sqrt{3\sqrt{x}-2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}{(x^{2}-4^{2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Using the above algebraic identity ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x})^{2}-(\sqrt{3\sqrt{x}-2})^{2}}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-(3\sqrt{x}-2)}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}+2}{\{(\sqrt{x})^{2}-2^{2}\}(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , here we <em>need</em> to <em>eliminate (√x-2)</em> from the denominator somehow , or the limit will again be <em>indeterminate </em>,so if you think <em>carefully</em> as <em>I thought</em> after <em>seeing the question</em> i.e what if we <em>add 4 and subtract 4</em> in <em>numerator</em> ? So let's try !

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2+4-4}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(x-4)+2+4-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , using the same above identity ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+6-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+3(2-\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , take minus sign common in <em>numerator</em> from 2nd term , so that we can <em>take (√x-2) common</em> from both terms

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)-3(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , take<em> (√x-2) common</em> in numerator ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)\{(\sqrt{x}+2)-3\}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Cancelling the <em>radical</em> that makes our <em>limit again and again</em> <em>indeterminate</em> ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\cancel{(\sqrt{x}-2)}\{(\sqrt{x}+2)-3\}}{\cancel{(\sqrt{x}-2)}(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}+2-3)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-1)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , <em>putting the limit ;</em>

{:\implies \quad \sf \dfrac{\sqrt{4}-1}{(\sqrt{4}+2)(4+4)(\sqrt{4}+\sqrt{3\sqrt{4}-2})}}

{:\implies \quad \sf \dfrac{2-1}{(2+2)(4+4)(2+\sqrt{3\times 2-2})}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{6-2})}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{4})}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(2+2)}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(4)}}

{:\implies \quad \sf \dfrac{1}{128}}

{:\implies \quad \bf \therefore \underline{\underline{\displaystyle \bf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}=\dfrac{1}{128}}}}

3 0
2 years ago
Read 2 more answers
What is 5/7*w=40 1/2
sukhopar [10]

Answer:

56 7/10

Step-by-step explanation:

convert 40 1/2 to a fraction 40 1/2 = 81/2

multiply both sides by  7/5

this gets w by itself on the left

81/2 x 7/5 = 567/10 = 56 7/10

7 0
3 years ago
Stan ran 4 7/10 miles, which was 1 3/10 fewer miles than Matt ran. Four students wrote and solved equations to find m, the numbe
nydimaria [60]

Answer: <u><em>Miles ran by Matt = 6 miles</em></u>

Step-by-step explanation: Miles ran by Stan = 4 7/10

i.e. 1 3/10 miles less than Matt.

∴ Miles ran by Matt = 4 7/10 + 1 3/10 = 60/10 = 6miles

∴<u><em> Miles ran by Matt = 6 miles</em></u>

<u><em /></u>

6 0
3 years ago
Read 2 more answers
Other questions:
  • Predict how much money can be saved without having a negative actual net income.
    7·1 answer
  • A druggist wants to reduce a 20-ounce quantity of a 7% solution of iodine to a 4% solution by adding pure alcohol. How much alco
    9·1 answer
  • To get to school each morning, Vanessa takes a horse 19.33 miles and a train 2.27 miles. In total the journey takes 26.48 minute
    10·2 answers
  • PLSS HELP
    7·1 answer
  • In a poll taken in December 2012, Gallup asked 1006 national adults whether they were baseball fans: 48% said they were. Almost
    14·1 answer
  • Help me its due today
    14·2 answers
  • Help me and fast plz
    12·1 answer
  • Work out the value of (2^3)^2
    9·1 answer
  • Plz help it’s due rn!!
    5·1 answer
  • Which equation has the same solution(s) as x2 + 6x + 1 = 0?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!