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3 years ago

11

PLSS HELPP!!! Thanks!

2 answers:

Airida [17]3 years ago

6 0

The Slope is 0 (there is no incline or decline therefore there is no slope)

adelina 88 [10]3 years ago

4 0

Answer:

0

Step-by-step explanation:

A vertical (straight up and down) line has an undefined slope

The y values change but the x values do not

m = (y2-y1)/ (x2-x1)

This makes the denominator 0

When we divide by 0, our answer is undefined

When we have a horizontal line, the slope is zero

The x values change but the y values do not

m = (y2-y1)/ (x2-x1)

This makes the numerator 0

The slope will be zero

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Someone please give me the answers for a through h

yulyashka [42]

A. Reflection across x. Left shift 2.
b. Right shift 3, up 6
c. Up 1
d. Reflect across x, right 5, down 2.
e. Down 4, vertical stretch 3.
f. Right 2, up 5, compression 2/5
g. Left 5
h. Reflect x down 4

3 0

3 years ago

Pls help me with this

3241004551 [841]

Answer: D 1 and 5/72 sq ft

8 0

3 years ago

10 points! doing brainliest! Do all 3<br><br><br>

faltersainse [42]

Answer:

12) $\frac{7}{3} = \frac{b}{18}$

$3b = 126$

$\frac{3b}{3} = \frac{126}{3}$

$b = 42$

13) $\frac{3.6}{m} = \frac{1.2}{3.6}$

$1.2m = 12.96$

$\frac{12m}{12} = \frac{12.96}{12}$

$m = 10.8$

14) $\frac{16}{10} = \frac{n + 2}{5}$

$80 = 10n + 20$

$80 - 20 = 10n +20 -20$

$60 = 10n$

$\frac{60}{10} = \frac{10n}{10}$

$n = 6$

7 0

3 years ago

Question is DOWN↓ PLS ANSWER I WILL MARK BRAINLIEST.....

Sonja [21]

Answer: Anita's

Step-by-step explanation:

Anita's shape would have the longest sides because she only has 4 total sides, while the others have 5, 6, and 10

This means that if the total perimeter for each of the shapes was 10 units,

Anita's shape would have side lengths of about 2.5 units while the others would have side lengths of 2, 1.6, and 1 units.

This means Anita's shape has the longest sides

5 0

2 years ago

F(x) = x^2+x-2/x^2-3x-4

Andrej [43]

i. Domain and Range

The given function is

$f(x)=\frac{x^2+x-2}{x^2-3x-4}$

The domain of this function is,

$x^2-3x-4\ne 0$

$(x-4)(x+1)\ne 0$

$x\ne4,xne -1$

The range refers to the y-values for which x is defined. x is defined for all values of y.

The range is all real numbers. See graph

ii. x-and-y-intercept

For x- intercept intercept we put $f(x)=0$

This implies that;

$\frac{x^2+x-2}{x^2-3x-4}=0$

This will give us

$x^2+x-2=0$

$\Rightarrow x^2+x-2=0$

$\Rightarrow x^2+2x--x-2=0$

$\Rightarrow x(x+2)-1(x+2)=0$

$\Rightarrow (x+2)(x-1)=0$

$\Rightarrow (x+2)=0,(x-1)=0$

$\Rightarrow x=-2,x=1$

The x-intercepts are $(-2,0),(1,0)$

For y-intercept, we put

$x=0$ to obtain;

$f(0)=\frac{0^2+0-2}{0^2-3(0)-4}$

$f(0)=\frac{1}{2}$

The y-intercept is

$(0,\frac{1}{2})$

iii. Horizontal asyptote

Since degree of the numerator and the denominator are the same, there is a horizontal asymptote

To find the horizontal asymptote.

We divide the leading coefficient of the numerator by the leading coefficient of the denominator.

The horizontal asymptote is $y=\frac{1}{1}=1$

iv. Vertical asymptote

To find the vertical asymptote, we equate the denominator to zero to get;

$x^2-3x-4=0$

This implies that;

$x^2+x-4x-4=0$

Split the middle term

$x(x+1)-4(x+1)=0$

Factor

$(x+1)(x-4)=0$

Factor further

$(x+1)=0,(x-4)=0$

$x=-1,x=4$

The vertical asymptotes are $x=-1,x=4$

8 0

3 years ago

Read 2 more answers