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3 years ago

10

A beetle is 1/5 inch wide and 2/5 inch long. How much greater is the beetle's length than its width?

1 answer:

Neporo4naja [7]3 years ago

6 0

Answer:

2 times as long

Step-by-step explanation:

2/5 divded by 1/5 equals two

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A line passes through point A (14,21). A second point on the line has an x-value that is 125% of the x-value of point A and a y-

seropon [69]

Answer:

The equation of the line in point-slope form is $y-21 = - \frac{3}{2}\cdot (x-14)$ .

Step-by-step explanation:

According to the statement, let $A(x,y) = (14,21)$ and $B(x,y) = (1.25\cdot x_{A},0.75\cdot y_{A})$ . The equation of the line in point-slope form is defined by the following formula:

$y-y_{A} = m\cdot (x-x_{A})$ (1)

Where:

$x_{A}$ , $y_{A}$ - Coordinates of the point A, dimensionless.

$m$ - Slope, dimensionless.

$x$ - Independent variable, dimensionless.

$y$ - Dependent variable, dimensionless.

In addition, the slope of the line is defined by:

$m = \frac{y_{B}-y_{A}}{x_{B}-x_{A}}$ (2)

If we know that $x_{A} = 14$ and $y_{A} = 21$ , then the equation of the line in point-slope form is:

$x_{B} = 1.25\cdot (14)$

$x_{B} = 17.5$

$y_{B} = 0.75\cdot (21)$

$y_{B} = 15.75$

From (2):

$m = \frac{15.75-21}{17.5-14}$

$m = -\frac{3}{2}$

By (1):

$y-21 = - \frac{3}{2}\cdot (x-14)$

The equation of the line in point-slope form is $y-21 = - \frac{3}{2}\cdot (x-14)$ .

5 0

3 years ago

NEED HELP PLEASE I BEG OF YOU

SVEN [57.7K]

Answer:

should be 40, because the top line is 40 and the line on the side is the same length

Step-by-step explanation:

5 0

3 years ago

Jim is experimenting with a new drawing program on his computer. He created quadrilateral TEAM with coordinates T(- 2, 3), E(- 5

Crazy boy [7]

Answer: yes Jim is correct

5 0

3 years ago

Can you help me with these two because I don't understand

OverLord2011 [107]

Answer: 1: 34 units 2: 24.5

Step-by-step explanation:

For figure 1, just split the area into squares and triangles and find the area of each. If you split figure 1 into 1 square and 3 triangles you get...

4x4= 16

(4x3)/2= 6

(4x4)/2= 8

(4x2)/2= 3

16+6+8+3 = 34

You do the same for figure 2, but use 4 triangles and 1 tiny square

1x1= 1

(4x3)/2= 6

(3x3)/2= 4.5

(4x4)/2= 8

(5x5)/2= 5

1+6+4.5+8+5= 24.5

You shouldn't need to worry about negative numbers because you are looking at the area

5 0

2 years ago

A. Use composition to prove whether or not the functions are inverses of each other. B. Express the domain of the compositions u

Kryger [21]

Given: $f(x) = \frac{1}{x-2}$

$g(x) = \frac{2x+1}{x}$

A.)Consider

$f(g(x))= f(\frac{2x+1}{x} )$

$f(\frac{2x+1}{x} )=\frac{1}{(\frac{2x+1}{x})-2}$

$f(\frac{2x+1}{x} )=\frac{1}{\frac{2x+1-2x}{x}}$

$f(\frac{2x+1}{x} )=\frac{x}{1}$

$f(\frac{2x+1}{x} )=1$

Also,

$g(f(x))= g(\frac{1}{x-2} )$

$g(\frac{1}{x-2} )= \frac{2(\frac{1}{x-2}) +1 }{\frac{1}{x-2}}$

$g(\frac{1}{x-2} )= \frac{\frac{2+x-2}{x-2} }{\frac{1}{x-2}}$

$g(\frac{1}{x-2} )= \frac{x }{1}$

$g(\frac{1}{x-2} )= x$

Since, $f(g(x))=g(f(x))=x$

Therefore, both functions are inverses of each other.

B.

For the Composition function $f(g(x)) = f(\frac{2x+1}{x} )=x$

Since, the function $f(g(x))$ is not defined for $x=0$ .

Therefore, the domain is $(-\infty,0)\cup(0,\infty)$

For the Composition function $g(f(x)) =g(\frac{1}{x-2} )=x$

Since, the function $g(f(x))$ is not defined for $x=2$ .

Therefore, the domain is $(-\infty,2)\cup(2,\infty)$

8 0

4 years ago