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3 years ago

12

What are the vertical asymptote(s) and hole(s) for the graph of y =

1 answer:

Murljashka [212]3 years ago

6 0

Answer:

asymptotes: x= -4,-2 and no holes

Step-by-step explanation:

First we need to factor the denominator of the function

y = (x-1) / (x2 + 6x + 8)

x2 + 6x + 8 = 0

delta = b2 - 4ac = 36 - 32 = 4

x1 = (-6 + 2) / 2 = -2

x2 = (-6 - 2) / 2 = -4

So we have that x2 + 6x + 8 = (x+4)(x+2)

So our function is:

y = (x-1) / (x+4)(x+2)

As there is no common expressions in the numerator and denominator, there are no holes.

The asymptotes are when the denominator is zero, so:

(x+4)(x+2) = 0

x = -4 or x = -2

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Convert 1 Iinto an improper fraction.

FromTheMoon [43]

Answer:

only mixed number can be changed into improper fraction according to my khowlage of grade7

Step-by-step explanation:

thank you

3 0

3 years ago

Evaluate the following pic

GREYUIT [131]

Answer:

1) $\sqrt{1225}+\sqrt{1024}=67$

2) $\sqrt[3]{-1331}=-11$

3) Evaluating $2:p :: p:8$ we get $p=\pm 4$

4) $x^3+y^2+z \ when \ x=3, y=-2, x=-6 \ we \ get \ \mathbf{25}$

5) $\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}=-6$

Step-by-step explanation:

1) $\sqrt{1225}+\sqrt{1024}$

Prime factors of 1225 : 5x5x7x7

Prime factors of 1024: 2x2x2x2x2x2x2x2x2x2

$\sqrt{1225}+\sqrt{1024}\\=\sqrt{5\times5\times7\times7}+\sqrt{2\times2\times2\times2\times2\times2\times2\times2\times2\times2}\\=\sqrt{5^2\times7^2}+\sqrt{2^2\times2^2\times2^2\times2^2\times2^2}\\=5\times7+(2\times2\times2\times2\times2)\\=35+32\\=67$

$\sqrt{1225}+\sqrt{1024}=67$

2) $\sqrt[3]{-1331}$

We know that $\sqrt[n]{-x}=-\sqrt[n]{x} \ ( \ if \ n \ is \ odd)$

Applying radical rule:

$\sqrt[3]{-1331}\\=-\sqrt[3]{1331} \\=-\sqrt[3]{11\times\11\times11}\\=-\sqrt[3]{11^3} \\Using \ \sqrt[n]{x^n}=x \\=-11$

So, $\sqrt[3]{-1331}=-11$

3) $2:p :: p:8$

It can be written as:

$p*p=2*8\\p^2=16\\Taking \ square \ root \ on \ both \ sides\\\sqrt{p^2}=\sqrt{16}\\p=\pm 4$

Evaluating $2:p :: p:8$ we get $p=\pm 4$

4) $x^3+y^2+z \ when \ x=3, y=-2, x=-6$

Put value of x, y and z in equation and solve:

$x^3+y^2+z \\=(3)^3+(-2)^2+(-6)\\=27+4-6\\=25$

So, $x^3+y^2+z \ when \ x=3, y=-2, x=-6 \ we \ get \ \mathbf{25}$

5) $\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}$

We know (-a)^n = (a)^n when n is even and (-a)^n = (-a)^n when n is odd

$\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}\\\\=\frac{1296\times-8\times 27}{46656}\\\\=\frac{-279936}{46656} \\\\=-6$

So, $\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}=-6$

8 0

3 years ago

WILL GIVE BRAINIEST FOR CORRECT ANSWER

lapo4ka [179]

If the center of dilation is A, this means that point A is invariant.

This means you can eliminate C and D.

Also, the scale factor is 3/4, which means that A'C' is more than half the length of AC.

This is best reflected by option B

7 0

2 years ago

Which of the following words is defined as “the study of hearing”? audiology audiologist audiometer audio

elena55 [62]

Audiology 100% hope this helps :)

4 0

3 years ago

(05.01)A scale drawing of a living room is shown below. The scale is 1 : 40.

matrenka [14]

Answer:

$A=266\frac{2}{3}\ ft^2$

Step-by-step explanation:

we know that

The scale is $\frac{1}{40}$

That means

1 unit in the drawing represent 40 units in the actual

or

1 inch in the drawing represent 40 inches in the actual

step 1

Find the actual dimensions of the living room

Multiply the dimensions of the living room in the drawing by 40

$Length:\ 6(40)=240\ in$

$Width:\ 4(40)=160\ in$

step 2

Convert inches to feet

Remember that

$1\ ft=12\ in$

To convert inches to feet , divide by 12

so

$Length:\ 240\ in=240/12=20\ ft$

$Width:\ 160\ in=160/12=\frac{40}{3}\ ft$

step 3

Find the area in square feet

Multiply the length by the width

$A=(20)(\frac{40}{3})=\frac{800}{3}\ ft^2$

Convert to mixed number

$A=\frac{800}{3}\ ft^2=\frac{798}{3}+\frac{2}{3}=266\frac{2}{3}\ ft^2$

4 0

3 years ago

Read 2 more answers