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3 years ago

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1 answer:

Aleks04 [339]3 years ago

7 0

Answer:

Figure A is a translation of Figure 1

hope it helps...

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A quadrilateral has congruent diagonals. It is probably a( n) ______.

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<span>A quadrilateral has congruent diagonals. It is probably a <u>Square</u></span>

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3 years ago

Whats the domain? someone pleaseee

ziro4ka [17]

-infinity, positive infinity

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3 years ago

Annie is framing a photo with a length of 6 inches and a width of 4 inches. The distance from the edge of the photo to the edge

Ede4ka [16]

Answer:

Part a) The quadratic function is $4x^{2} +20x-39=0$

Part b) The value of x is $1.5\ in$

Part c) The photo and frame together are $7\ in$ wide

Step-by-step explanation:

Part a) Write a quadratic function to find the distance from the edge of the photo to the edge of the frame

Let

x----> the distance from the edge of the photo to the edge of the frame

we know that

$(6+2x)(4+2x)=63\\24+12x+8x+4x^{2}=63\\ 4x^{2} +20x+24-63=0\\4x^{2} +20x-39=0$

Part b) What is the value of x?

Solve the quadratic equation $4x^{2} +20x-39=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem

we have

$4x^{2} +20x-39=0$

so

$a=4\\b=20\\c=-39$

substitute in the formula

$x=\frac{-20(+/-)\sqrt{20^{2}-4(4)(-39)}} {2(4)}$

$x=\frac{-20(+/-)\sqrt{1,024}} {8}$

$x=\frac{-20(+/-)32} {8}$

$x=\frac{-20(+)32} {8}=1.5\ in$ -----> the solution

$x=\frac{-20(-)32} {8}=-6.5\ in$

Part c) How wide are the photo and frame together?

$(4+2x)=4+2(1.5)=7\ in$

5 0

3 years ago

The quotient of 24 and a number is 6.

Flauer [41]

The answer will be: 4

24 / 6 = 4

6 0

3 years ago

Read 2 more answers

Given the parabola below, find the endpoints of the latus rectum. (x-2)^2=-20(y+2)

Shtirlitz [24]

Answer:

The endpoints of the latus rectum are $(12, -7)$ and $(-8, -7)$ .

Step-by-step explanation:

A parabola with vertex at point $C(x, y) = (h,k)$ and whose axis of symmetry is parallel to the y-axis is defined by the following formula:

$(x-h)^{2} = 4\cdot p \cdot (y-k)$ (1)

Where:

$y$ - Independent variable.

$x$ - Dependent variable.

$p$ - Distance from vertex to the focus.

$h$ , $k$ - Coordinates of the vertex.

The coordinates of the focus are represented by:

$F(x,y) = (h, k+p)$ (2)

The <em>latus rectum</em> is a line segment parallel to the x-axis which contains the focus. If we know that $h = 2$ , $k = -2$ and $p = -5$ , then the latus rectum is between the following endpoints:

By (2):

$F(x,y) = (2, -2-5)$

$F(x,y) = (2,-7)$

By (1):

$(x-2)^{2} = -20\cdot (-7+2)$

$(x-2)^{2} = 100$

$x - 2 = \pm 10$

There are two solutions:

$x_{1} = 2 + 10$

$x_{1} = 12$

$x_{2} = 2-10$

$x_{2} = -8$

Hence, the endpoints of the latus rectum are $(12, -7)$ and $(-8, -7)$ .

4 0

2 years ago