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natita [175]
3 years ago
5

Where are the astmptotes for the following function located f(x)=7/x^2-2x-24​

Mathematics
1 answer:
ira [324]3 years ago
6 0

Answer:

Hence, the asymptotes of f(x) located at x=6 and x=-4.

Step-by-step explanation:

Since we have given that

f(x)= \frac{7}{x^2-2x-24}

We need to find the asymptotes for the above function:

Asymptotes occur when denominator becomes zero.

x^2-2x-24=0\\\\x^2-6x+4x-24=0\\\\x(x-6)+4(x-6)=0\\\\(x-6)(x+4)=0\\\\x=6,-4

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(Marking Brainliest) What is the vertex form of the equation?<br><br>y = x^2 - 4x + 7​
AysviL [449]

Step-by-step explanation:

y = x² - 4x + 7

the general vertex form is

y = m(x-h)² + k

to bring the part "x² -4x" to an expression of (ax + b)² we need to add 4, as "x² - 4x + 4" = (x - 2)².

and since we add 4 there, we need to subtract 4 overall again to keep the value of the expression the same :

y = x² - 4x + 4 + 7 - 4 = (x - 2)² + 7 - 4 = (x - 2)² + 3

and so, that is the vertex form :

y = (x - 2)² + 3

4 0
3 years ago
Simplify the expression by arranging the steps in sequence based on the order of operations 1/6(30x+24)-1/7(63-7×)
alex41 [277]

Answer:

6x-5

Step-by-step explanation:

We apply distributive property to get rid of the first parenthesis, by multiplying each term inside by the factor 1/6:

\frac{1}{6} (30x+24)-\frac{1}{7} (63-7x)\\\frac{30}{6} x+\frac{24}{6} -\frac{1}{7} (63-7x)\\5x+4-\frac{1}{7} (63-7x)

now we get rid of the second parenthesis by multiplying the factor "-\frac{1}{7}" times each of the terms inside the parenthesis (notice here that this factor is negative , so it will flip the signs of the terms it multiplies):

5x+4-\frac{1}{7} (63-7x)\\5x+4-\frac{63}{7} +\frac{7}{7} x\\5x+4-9+x

Next we combine like terms:

5x+4-9+x\\5x+x+4-9\\6x-5

7 0
3 years ago
Jessa walked 3 miles.
valentina_108 [34]

Answer:

190080 in.

Step-by-step explanation:

5280 x 3 = 15840

15840 x 12 = 190080

7 0
2 years ago
Read 2 more answers
The picture below shows Damarian standing at a corner of a field that is in the shape of a rhombus. His friend is standing in th
Nitella [24]

Answer:

The distance between Damarian and his friend is approximately 127 feet

Step-by-step explanation:

  • The four sides of the rhombus are equal
  • The diagonals of the rhombus bisects each other and perpendicular to each other

<em>Look to the attached figure</em>

∵ ADCF is a rhombus

∵ The distance of each side of the field is 90 feet

- The four sides of the rhombus are equal

∴ AD = DC = CF = FA = 90

∵ AC and DF are its diagonals

∴ AC ⊥ DF and bisects each other

∵ D is the position of Damarian

∵ F is the position of his friend

∵ The distance between the other two corners of the field is

   approximately 128 feet

∴ AC = 128 feet

∵ AC and DF bisects each other at point I

∴ AI = CI

- That means each part is one-half AC

∴ AI = CI = 128 ÷ 2 = 64 feet

<em>Now let us use any right triangle of the four right triangles in the figure to find the length of one-half the other diagonal</em>

In Δ FIC

∵ ∠FIC is a right angle

∵ FC = 90 ⇒ The hypotenuse of the Δ

∵ CI = 64 ⇒ one leg of the Δ

- By using Pythagoras Theorem

∴ (FC²) = (IF)² + (CI)²

- Substitute the values if CI and FC in the formula of Pythagoras

∵ (90)² = (IF)² + (64)²

∴ 8100 = (IF)² + 4096

- Subtract 4096 from both sides

∴ 4004 = (IF)²

- Take √  for both sides

∴ 63.277 = IF

∵ DI = IF = \frac{1}{2} DF

- Multiply DI by 2 to find DF

∴ DF = 126.554

- Round it the the nearest whole number

∴ DF = 127 feet

DF represents the distance between Damarian and his friend

The distance between Damarian and his friend is approximately 127 feet

8 0
3 years ago
What is the value of the expression?
nadya68 [22]
B use the app Photomath it helps you with this kinda questions
8 0
3 years ago
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