Ask question

Ask question

All categories

VashaNatasha [74]

3 years ago

10

Can you please help me i give extra if correct!

1 answer:

vazorg [7]3 years ago

4 0

Answer:

Is there a multiple choice?

Step-by-step explanation:

You might be interested in

Sharon has a hula hoop with a circumference of 88 inches. What is the approximate diameter of her hula hoop?

timurjin [86]

Answer:

28.0

Step-by-step explanation:

The circumference is diameter*3.14 (pi). So for this you divide to get the diameter. Hope this helps! :)\

88/3.14=28.0

3 0

3 years ago

Which is the graph of the sequence defined by the function f(x + 1) = 2/3 f(x) if the initial value of the sequence is 108?

san4es73 [151]

Answer:

The fourth graph.

Step-by-step explanation:

We have f(x + 1) so the x values will be increased by 1.

first value = 108 (when x = 1)

next value = 108 * 2/3 = 72 (when x = 2)

next = 72 * 2/ 3= 48 (when x = 3)

next = 48 * 2/3 = 32 (when x = 4)

5 0

3 years ago

Read 2 more answers

I don't know how to do this. I have the formula but I'm confused. Help?

Rom4ik [11]

For the lateral area, don't include the base area. For the surface area, you should include it.

3 0

3 years ago

A box of Munchkins contains chocolate and glazed donut holes. If Jacob ate 2 chocolate

neonofarm [45]

Answer:

24 munchkins.

Step-by-step explanation:

Let C be the number of chocolate and D be number of glazed donut holes in the original box.

We are told if Jacob ate 2 chocolate munchkins, then 1/11 of the remaining Munchkins would be chocolate. We can represent this information as:

$C-2=\frac{1}{11}*(C+D-2)...(1)$

We are also told if he instead added 4 glazed Munchkins to the original box, 1/7 of the Munchkins would be chocolate. We can represent this information as:

$C=\frac{1}{7}*(C+D+4)...(2)$

Upon substituting C's value from equation (2) in equation (1) we will get,

$\frac{1}{7}*(C+D+4)-2=\frac{1}{11}*(C+D-2)$

Let us have a common denominator on right side of equation.

$\frac{1}{7}*(C+D+4)-\frac{7*2}{7}=\frac{1}{11}*(C+D-2)$

$\frac{C+D+4-14}{7}=\frac{1}{11}*(C+D-2)$

Multiplying both sides of our equation by 7, we will get,

$7*\frac{C+D-10}{7}=7*\frac{1}{11}*(C+D-2)$

$C+D-10=\frac{7}{11}*(C+D-2)$

Multiplying both sides of our equation by 11, we will get,

$11*(C+D-10)=11*\frac{7}{11}*(C+D-2)$

$11*(C+D-10)=7*(C+D-2)$

$11C+11D-110=7C+7D-14$

$11C-7C+11D-7D=-14+110$

$4C+4D=96$

$4(C+D)=96$

$(C+D)=\frac{96}{4}$

$(C+D)=24$

Therefore, the total number of Munchkins in original box is 24.

3 0

3 years ago

Solve the following equation by factoring:9x^2-3x-2=0

olya-2409 [2.1K]

Answer:

The two roots of the quadratic equation are

$x_1= - \frac{1}{3} \text{ and } x_2= \frac{2}{3}$

Step-by-step explanation:

Original quadratic equation is $9x^{2}-3x-2=0$

Divide both sides by 9:

$x^{2} - \frac{x}{3} - \frac{2}{9}=0$

Add $\frac{2}{9}$ to both sides to get rid of the constant on the LHS

$x^{2} - \frac{x}{3} - \frac{2}{9}+\frac{2}{9}=\frac{2}{9}$ ==> $x^{2} - \frac{x}{3}=\frac{2}{9}$

Add $\frac{1}{36}$ to both sides

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{2}{9} +\frac{1}{36}$

This simplifies to

$x^{2} - \frac{x}{3}+\frac{1}{36}=\frac{1}{4}$

Noting that (a + b)² = a² + 2ab + b²

If we set a = x and b = $\frac{1}{6}\right)$ we can see that

$\left(x - \frac{1}{6}\right)^2$ = $x^2 - 2.x. (-\frac{1}{6}) + \frac{1}{36} = x^{2} - \frac{x}{3}+\frac{1}{36}$

So

$\left(x - \frac{1}{6}\right)^2=\frac{1}{4}$

Taking square roots on both sides

$\left(x - \frac{1}{6}\right)^2= \pm\frac{1}{4}$

So the two roots or solutions of the equation are

$x - \frac{1}{6}=-\sqrt{\frac{1}{4}}$ and $x - \frac{1}{6}=\sqrt{\frac{1}{4}}$

$\sqrt{\frac{1}{4}} = \frac{1}{2}$

So the two roots are

$x_1=\frac{1}{6} - \frac{1}{2} = -\frac{1}{3}$

and

$x_2=\frac{1}{6} + \frac{1}{2} = \frac{2}{3}$

7 0

2 years ago