All categories

2 years ago

Find the slope y = -1+ 3x?

Mathematics

2 answers:

olga nikolaevna [1]2 years ago

4 0

Answer:

the slope is 3

Step-by-step explanation:

Inga [223]2 years ago

4 0

Answer:

-1 because the slope is the m.

Step-by-step explanation:

y=mx+b

You might be interested in

Help please just answer would be fine

scZoUnD [109]

The amount of water decreases $16\frac{1}{2}$ pints in 12 weeks.

Solution:

The amount of water decreases in 7 weeks = $9\frac{5}{8}$ pints

The amount of water decreases each week is same.

Let us first change the mixed fraction into improper fraction.

$9\frac{5}{8}$ pints = $\frac{(9\times8)+5}{8}$

$=\frac{77}{8}$ pints

The amount of water decreases in 1 week

$$=\frac{\frac{77}{8} }{7}$

$=\frac{77}{8\times7}$

$=\frac{11}{8}$

So, the amount of water decreases in 1 week is $\frac{11}{8}$ pints.

The amount of water decreases in 12 week

$=\frac{11}{8}\times12$

$=\frac{132}{8}$

$=16\frac{4}{8}$

$=16\frac{1}{2}$ pints

Hence the amount of water decreases $16\frac{1}{2}$ pints in 12 weeks.

4 0

3 years ago

The ratio of Jared’s stamps to Pablo’s stamps is 4:7. if Jared has 36 stamps how many stamps does Paulo have?

chubhunter [2.5K]

36/4=9
9x7=63
Paulo has 63 stamps

5 0

2 years ago

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

$\therefore \: x = 5$