Evaluate 4x(4+ (2(6) x2(-3))) Please help!!!!

Zoe had 3 1 /8 feet of orange ribbon. She used some ribbon to make a bow for a gift now she has 1 3 /8 feet of orange ribbon lef

Nastasia [14]

Answer:

Orange ribbon Zoe use is $1\frac{3}{4} \ feet.$

Step-by-step explanation:

Given:

Zoe had 3 1 /8 feet of orange ribbon. She used some ribbon to make a bow for a gift now she has 1 3 /8 feet of orange ribbon left.

Now, to find the orange ribbon Zoe use.

Let the orange ribbon Zoe use be $x.$

Total orange ribbon Zoe had = $3\frac{1}{8} = \frac{25}{8} \ feet.$

Ribbon Zoe had left = $1\frac{3}{8} =\frac{11}{8} \ feet.$

Now, to get the orange ribbon Zoe use:

$\frac{25}{8} -x=\frac{11}{8}$

Subtracting both sides by $\frac{11}{8}$ we get:

$\frac{25}{8} -x-\frac{11}{8} =0$

Adding both sides by $x$ we get:

$\frac{25}{8} -\frac{11}{8} =x$

$\frac{25-11}{8} =x$

$\frac{14}{8}=x$

$x=\frac{14}{8}$

$x=\frac{7}{4}$

$x=1\frac{3}{4}\ feet.$

Therefore, orange ribbon Zoe use is $1\frac{3}{4} \ feet.$

8 0

3 years ago

In a certain class of students, there are 11 boys from Wilmette, 7 girls from Wilmette, 5 boys from Winnetka, 4 girls from

attashe74 [19]

Answer:

.571

Step-by-step explanation:

We need to find the total number of boys

11+5+4 = 20

And the total number of students

11+7+5+4+4+4 = 35

The Probability that the student is a boy is boy/total

P (boy) = boy/total

= 20/35

=4/7

=.571428571

to 3 decimal places

=.571

6 0

3 years ago

The equation 7^2=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

jeyben [28]

Answer:

The period of Y increases by a factor of $k^ {3/2}$ with respect to the period of X

Step-by-step explanation:

The equation $T ^ 2 = a ^ 3$ shows the relationship between the orbital period of a planet, T, and the average distance from the planet to the sun, A, in astronomical units, AU. If planet Y is k times the average distance from the sun as planet X, at what factor does the orbital period increase?

For the planet Y:

$T_y ^ 2 = a_y ^ 3$

For planet X:

$T_x ^ 2 = a_x ^ 3$

To know the factor of aumeto we compared $T_x$ with $T_y$

We know that the distance "a" from planet Y is k times larger than the distance from planet X to the sun. So:

$a_y ^ 3 = (a_xk) ^ 3$

So

$\frac{T_y ^ 2}{T_x ^ 2}=\frac{a_y ^ 3}{a_x^ 3}\\\\\frac{T_y ^ 2}{T_x ^ 2}=\frac{(a_xk)^3}{a_x ^ 3}\\\\\frac{T_y^ 2}{T_x^ 2}=\frac{k ^ {3}a_{x}^ 3}{a_{x}^ 3}\\\\\frac{T_{y}^ 2}{T_{x}^ 2}=k ^ 3\\\\T_{y}^ 2 = T_{x}^{2}k^{3}\\\\T_{y} =k^{\frac{3}{2}}T_x$

Then, the period of Y increases by a factor of $k^ {3/2}$ with respect to the period of X

4 0

3 years ago

Diouna has $35,000 worth of rental car insurance. She rented a Lexus over the weekend and got into a car accident causing the ca

Dimas [21]

Answer:

she will be responsible with a big bill to pay that's for sure

3 0

3 years ago

Square root of both sides of (x + 9)2 = 25?

anygoal [31]

Answer: C x+9 = 5

Step-by-step explanation:

6 0

3 years ago