All categories

2 years ago

Please explain!! 20 points and I will mark brainliest

Mathematics

1 answer:

nika2105 [10]2 years ago

8 0

No, her current average is 225. She would need to score 350 points to get a 250.

You might be interested in

postnew [5]

12-3=9 they switch up and it still equals the same thing

5 0

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

2x + (3)(x) + (3)(4)

S_A_V [24]

Answer:

if you were looking for the solution i think it is Solution

5 + 1 2

Step-by-step explanation:

4 0

1 year ago

Read 2 more answers

Use the given minimum and maximum data entries, and the number of classes, to find the class width, the lower class limits,

Gennadij [26K]

Using proportions and the information given, it is found that:

The class width is of 14.375.

The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.

The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.

-------------------------

Minimum value is 19.
Maximum value is of 134.
There are 8 classes.
The classes are all of equal width, thus the width is of:

$W = \frac{134 - 19}{8} = 14.375$

-------------------------

The intervals will be of:

19 - 33.375

33.375 - 47.750

47.750 - 62.125

62.125 - 76.500

76.500 - 90.875

90.875 - 105.250

105.250 - 119.625

119.625 - 134.

The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.

The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.

A similar problem is given at brainly.com/question/16631975

6 0

2 years ago

Can someone please solve for x

RoseWind [281]

Answer:

x = -11

Step-by-step explanation:

2x + 23 + 9 = x + 21

Combine common factors

2x + 32 = x + 21

Subtract x from both sides

x + 32 = 21

Subtract 32 from both sides

x = -11

5 0

2 years ago