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3 years ago

What does it mean when a figure is dilated by a scale factor of 1/4

Mathematics

2 answers:

Harlamova29_29 [7]3 years ago

7 0

When a figure is dilated by a scale factor of 1/4 it means it is 1/4 the original size

docker41 [41]3 years ago

3 0

It means the figure is 1/4 the size of the original figure

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HELP QUICK PLEASE!!!!

Simora [160]

Answer:

Step-by-step explanation:

Given the expression;

$3sin^2x + 4sinx - 4 = 0$

Let P = sinx

The expression becomes;

3P²+4P - 4 = 0

Factorize

3P²+6P-2P - 4 = 0

3P(P+2)-2(P+2) = 0

3P-2 = 0 and P+2 = 0

P = 2/3 and -2

When P = 2/3

sinx = 2/3

x = arcsin 2/3

x = arcsin 0.6667

x = 41.8 degrees

Also if P = -2

sinx = -2

x = arcsin (-2)

x will not exist in this case

To get other values of x

sin is positive in the second quadrant

x = 180 - 41.8

x = 138.2°

x = 360+41.8

x = 401.8°

<em>Hence the values of x within the interval are 41.8°, 138.2° and 401.8°</em>

7 0

3 years ago

Which rational number could be graphed between -3 and -4

Snezhnost [94]

Answer:

-3.5

Step-by-step explanation:

An example is -3.5.

3 0

4 years ago

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Pls help ill give brainliest !!

AnnyKZ [126]

Answer:

Step-by-step explanation:

4 0

3 years ago

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Write an equation for an ellipse centered at the origin, which has foci at (\pm5,0)(±5,0)(, plus minus, 5, comma, 0, )and vertic

Virty [35]

Answer:

The equation of ellipse is

$\frac{x^2}{41}+\frac{y^2}{16}=1$

Step-by-step explanation:

The equation of an ellipse is

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

where(h,k) is the center and c is distance from the center to the foci is given by $a^2-b^2=c^2$ . a is the distance from the center to the vertices and b is the distance from the center to the co-vertices.

The center of the ellipse is the mid-point of the vertices.

The mid point of the vertices $(\pm\sqrt{41},0)$ is

= $(\frac{\sqrt{41}+(-\sqr{41})}2,\frac{0+0}2)$

=(0,0)

a is the distance between the center and the vertices.

So, $a=\sqrt{(0-\sqrt{41})^2+(0-0)^2}$

$=\sqrt{41}$

c is the distance between the center and the foci.

So, $c=\sqrt{(0-5)^2+(0-0)^2$

$a^2-b^2=c^2$

$\Rightarrow (\sqrt41)^2-b^2=5^2$

$\Rightarrow b^2=(\sqrt41)^2- 5^2$

$\Rightarrow b^2=41-25$

$\Rightarrow b^2=16$

The equation of ellipse is

$\frac{(x-0)^2}{(\sqrt{41})^2}+\frac{(y-0)^2}{16}=1$

$\Rightarrow \frac{x^2}{41}+\frac{y^2}{16}=1$

7 0

4 years ago

Find the least length of a rope which can be cut into whole number of pieces of lengths 45cm, 75cm and 81 cm

choli [55]

The least is 45 bc it is less than the other 2

5 0

3 years ago