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3 years ago

If square HIJK is dilated by a scale factor of

Mathematics

2 answers:

WITCHER [35]3 years ago

3 0

C. point K is (3,-3) and the dilation moves from 3 to 1 so you divide both X and Y coordinates by 3 to get (1,-1)
hope this helps.

Sav [38]3 years ago

3 0

Answer:

C) (1, -1)

Step-by-step explanation:

When a figure is dilated, the coordinates of each point are multiplied by the scale factor. The scale factor in this problem is 1/3; this means that every point gets multiplied by 1/3.

The coordinates of K are (3, -3). Applying the dilation, we have

1/3(3, -3) = (1/3*3, 1/3*-3) = (1, -1)

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If 4x^2 - 100 = 0 then the roots of the equation are

iVinArrow [24]

4x² - 100 = 0
 + 100 + 100
 4x² = 100
 4 4
 x² = 25
 x = +5
4(5)² - 100 = 0
4(25) - 100 = 0
100 - 100 = 0
 0 = 0

4(-5)² - 100 = 0
4(25) - 100 = 0
 100 - 100 = 0
 0 = 0

The roots of the equation is positive or negative 5.

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2 years ago

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What is 2- 1/4 Write your answer as a fraction Meh will mark you

Arlecino [84]

Answer:

1 3/4

Step-by-step explanation:

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2 years ago

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THE RIGHT ANSWERRRRRR

Mkey [24]

Area of semicircle = 1/2 the area of a circle:
$A_{sc}=\frac{1}{2} \pi r^{2}$
radius is half the diameter of 6ft.
$A_{sc} = \frac{1}{2} \pi (3^{2} )=14.13 ft^{2} .$
Area of the rectangle:
$A_{r} =lw=15*6=90 ft^{2} .$
Add the two together:
90 + 14.13 = 104.13
Round to the nearest tenth (one decimal place)
Area = 104.1 square feet.

4 0

3 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

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

We split the middle term,

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

We factor to get,

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

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

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

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

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3 years ago

Write the following fraction as a decimal. A. 1.10 B. 10.1 C. 0.1 D. 0.01

skelet666 [1.2K]

A) 1.1/1=11/10

B)10.1/1=101/10

C).1/1=1/10

D).01/1=1/100

you put the number you have to multiply the decimal with to get a whole n.o as the decimal

Hope you found this answer useful

7 0

2 years ago

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