Thank y’all for helping please help !!

Lyn puts dvds on shelves. She has 236 dvds.each full shelf has 28 dvds.how many full shelves of dvds does Lyn have?

cluponka [151]

Answer:

8 shelves

Step-by-step explanation:

As 28 multiple by 2 is 224 and 28*9 is 252 so she will have 8 full shelves

3 0

3 years ago

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Please help with math show your work

zzz [600]

(5)(−3)−(4)(−3)−3−(3)(−3)=−15−(4)(−3)−3−(3)(−3)=−15−(−12)−3−(3)(−3)=−3−3−(3)(−3)=−6−(3)(−3)=−6−(−9)=3

5 0

3 years ago

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What is the center and radius of the circle with equation x2 + y2 - 4x + 22y + 61 = 0?

lesantik [10]

Hello,

$x^2+y^2-4x+22y+61=0\\\\
(x^2-4x+4) +(y^2+22y+121)-121-4+61=0\\\\
(x-2)^2+(y+11)^2=64


$

Center is (2,-11) and radius is 8

6 0

4 years ago

What’s the triangle congruence shortcut

mina [271]

Answer:

D. All those resources can be found in Africa

7 0

3 years ago

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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

$\therefore \: x = 5$

7 0

4 years ago