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3 years ago

Bruce evaluates the expression on a math quiz, 8.8 ÷ 2.7 and gets an answer of 4.14.

Mathematics

2 answers:

jolli1 [7]3 years ago

6 0

Answer:

the answer is D

Step-by-step explanation:

ella [17]3 years ago

4 0

The correct answer is choice d.....

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How can you determine the number of solutions for an equation

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The way to determine the number of solutions is to complete the problem in multiple different ways.

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2 years ago

Write an expression for "8 times z."

garri49 [273]

Answer: 8z 8(z)

Explanation: well 8 times z is simply 8(z)I don’t really think there is another way

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2 years ago

What are the coordinates of vertex A of square ABCD?

bonufazy [111]

ABCD is a square . if the coordinates of three of its vertices are A(-1,2a), B(a,2a), C(a,0), find the coordinates of D. BC is obviously 2a (it's a vertical line) then AB must be 2a (notice AB is horizontal) so BC = AC because it is a square 2a = a+1 a=1

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2 years ago

Read 2 more answers

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

3 years ago

5 Consider this system of equations:<br> 5x + y = -2<br> 2x - 2y=4

Ilia_Sergeevich [38]

Answer:

x = 0

y = -2

Step-by-step explanation:

The given system of equation are:

5x + y = -2

2x - 2y = 4

The problem here is to find x and y from the expression

5x + y = -2 ----- i

2x - 2y = 4 ---- ii

So;

i x 2 : 10x + 2y = -4 ---- iii

ii x 5 : 10x - 10y = 20 ----- iv

iii - iv; 12y = -24

y = -2

Now find x;

5x + y = -2

5x - 2 = -2

x = 0

7 0

2 years ago