How many petals are present when there are 12 leaves

A company ordered paper supplies. They bought five times white

Law Incorporation [45]

Answer:

they ordered 10 sheets of green paper

Step-by-step explanation:

they would have to be equal since there was five times white as green and green cost five cents

given this you would split it 50 50

50 divided by 5 is 10 and 50 divided by 1 is 50

hope this helps and if it is right could I please have a brainliest?

7 0

3 years ago

Clare bought a new pair of shorts for $ 13.75. This was 25% off the regular price. What was the original price? ( Round your ans

Inessa05 [86]

Answer:

18.3

Step-by-step explanation: Here i'll give you my notes for finding original price

1. make the percent into a decimal

2. subtract the decimal by 1

3. divide the number you just got and the decimal you got from the first step

7 0

2 years ago

Which of the following

Damm [24]

Answer:

A: represents a function

Explanation:

Every other option, A, has two y for one x.

It can be a function if there is 1x for 1y.

If this was helpful, please mark brainliest. Have a beautiful day!

4 0

3 years ago

Read 2 more answers

Mario made a table with the rules below

Lady_Fox [76]

It will be option A.
x - coordinate is 1. If we multiply 3 with 1, it will be 3. And if we multiply 3 with 3, it will be 9.
y- coordinate is 1. If we add 3 with 1, it will be 4. And adding 3 with 4, we get 7.

Answer:

a- (1,1) (3,4) (9,7)

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0

3 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

$\therefore \: x = 5$

7 0

4 years ago