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3 years ago

8

State the missing reasons in this flow proof.

1 answer:

posledela3 years ago

8 0

Is there a photo for this problem? I solved what i coulda. givenb. givenc. def. of supplementary angles (i think)d.photo to support this?
e.

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<img src="https://tex.z-dn.net/?f=x%5E2-11x%2B24%3D0" id="TexFormula1" title="x^2-11x+24=0" alt="x^2-11x+24=0" align="absmiddle"

Mashcka [7]

Your answer is x=-24/11

8 0

3 years ago

how to use Lcd in this problem? [a-2. -_1_ =_3_]find Lcd [ a+3 1 a-2]. (a+3)(a+2). multiply all neumerator to (LCD) a-2 - 1= 3 _

san4es73 [151]

It's not entirely clear to me what you're trying to solve, but it looks like the initial equation is

$\dfrac{a-2}{a+3} -1 = \dfrac3{a+2}$

First convert each term into a fraction with the same (i.e. the least common) denominator. The first term needs to be multiplied by <em>a</em> + 2; the second term by (<em>a</em> + 3) (<em>a</em> + 2); and the third term by <em>a</em> + 3 :

$\dfrac{a-2}{a+3}\cdot\dfrac{a+2}{a+2} -1\cdot\dfrac{(a+3)(a+2)}{(a+3)(a+2)} = \dfrac3{a+2}\cdot\dfrac{a+3}{a+3} \\\\ \dfrac{(a-2)(a+2)}{(a+3)(a+2)} - \dfrac{(a+3)(a+2)}{(a+3)(a+2)} = \dfrac{3(a+3)}{(a+3)(a+2)}$

Now that everything has the same denominator, we can combine the fractions into one. Move every term to one side and join the numerators:

$\dfrac{(a-2)(a+2)-(a+3)(a+2)-3(a+3)}{(a+3)(a+2)} = 0$

Simplify the numerator:

$\dfrac{(a^2-4)-(a^2+5a+6)-(3a+9)}{(a+3)(a+2)} = 0 \\\\ \dfrac{-8a-19}{(a+3)(a+2)} = 0$

If neither <em>a</em> = -3 nor <em>a</em> = -2, we can ignore the denominator:

$-8a-19 = 0$

Solve for <em>a</em> :

$-8a = 19 \\\\ \boxed{a = -\dfrac{19}8}$

4 0

3 years ago

The left and right page numbers of an open book are two consecutive integers whose sum is 347. Find these page numbers.

azamat

Let the first page = x

The second page would be x + 1

Add them together to equal 347

X + x + 1 = 347

Combine like terms:

2x + 1 = 347

Subtract 1 from both sides

2x = 346

Divide both sides by 2:

X = 173

The smaller page number is 173

The larger page number is 174

7 0

3 years ago

Read 2 more answers

Solve for w in the proportion.

Svet_ta [14]

Answer:

w=40

Step-by-step explanation:

5 0

3 years ago

A picture inside a frame is 2 inches longer than it is wide. The picture is in a frame that has a width of 3 inches on each side

geniusboy [140]

Answer:

The dimensions of the frame are 15 in x 13 in

Step-by-step explanation:

Let

x ----> the length of the picture

y ----> the width of the picture

we know that

$x=2+y$ -----> equation A

The area of the picture, including the frame is

$A=(x+6)(y+6)$

$A=195\ in^2$

so

$195=(x+6)(y+6)$ ----> equation B

substitute equation A in equation B

$195=(2+y+6)(y+6)$

solve for y

$195=(y+8)(y+6)\\y^2+6y+8y+48=195\\\\y^2+14y-147=0$

solve the quadratic equation by graphing

The solution is y=7 in

see the attached figure

<em>Find the value of x</em>

$x=2+7=9\ in$

Find the dimensions of the frame

$x+6=9+6=15\ in$

$y+6=7+6=13\ in$

therefore

The dimensions of the frame are 15 in x 13 in

7 0

3 years ago