Ask question

Ask question

All categories

anzhelika [568]

3 years ago

6

A cone shaped cup has a radius of 4 inches and a height of 6 inches. Which measurement is the closest to the volume in cubic inc

hes for the cup? Use 3.14 for π.
25.12 in³

12.28 in³

100.48 in³

6.14 in³

1 answer:

mihalych1998 [28]3 years ago

4 0

7488in3 in direct form of the eqation is the biggest answer but put 554in3 if in the parlogical form

You might be interested in

Can someone help me please

alekssr [168]

Use the formula x+x+1+x+2 or 3x+3=170 to find x. x is the first number. Then add x+1 to find the second number. Then add x+2 to find the third final number.
x will equal 167/3 which is a decimal, so just plug it in to the 3 equations I mentioned above to find your 3 consecutive numbers. More questions? Just ask me!

4 0

3 years ago

Read 2 more answers

Rachel bought 2 pounds of apples and 3 pounds of peaches for a total of $20.45. The apples and peaches cost the same amount per

Likurg_2 [28]

Each pound would cost at least 4.09$

8 0

3 years ago

Billy has a gallon of paint. He us going to pour it into a paint tray that measures 10 inches wide, 14 inches long and 4cm deep.

erica [24]

Answer:

Tray will overflow with $10.53\ \text{inch}^3$ of paint.

Step-by-step explanation:

Dimensions the tray is 10 inch by 14 inch by 4 cm

$1\ \text{cm}=\dfrac{1}{2.54}\ \text{inch}$

$4\ \text{cm}=\dfrac{4}{2.54}\ \text{inch}$

Volume the tray can hold is

$10\times 14\times \dfrac{4}{2.54}\ \text{in}^3$

$1\ \text{inch}^3=\dfrac{1}{231}\ \text{gallon}$

The volume of paint Billy has is $1\ \text{gallon}$

Difference in the volume of paint and volume of tray in cubic inches is

$\left(1-\dfrac{10\times14\times\dfrac{4}{2.54}}{231}\right)231=10.53\ \text{inch}^3$

The tray will overflow with $10.53\ \text{inch}^3$ of paint.

3 0

2 years ago

Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.

Let's start with setting the $y$ 's equal to find those $x$ 's for which the $y$ 's are the same.

$\log(x)=\frac{1}{2}\log(x+1)$

By power rule:

$\log(x)=\log((x+1)^\frac{1}{2})$

Since $\log(u)=\log(v)$ implies $u=v$ :

$x=(x+1)^\frac{1}{2}$

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

This is a quadratic equation.

Subtract $(x+1)$ on both sides:

$x^2-(x+1)=0$

$x^2-x-1=0$

Comparing this to $ax^2+bx+c=0$ we see the following:

$a=1$

$b=-1$

$c=-1$

Let's plug them into the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$x=\frac{1 \pm \sqrt{1+4}}{2}$

$x=\frac{1 \pm \sqrt{5}}{2}$

So we have the solutions to the quadratic equation are:

$x=\frac{1+\sqrt{5}}{2}$ or $x=\frac{1-\sqrt{5}}{2}$ .

The second solution definitely gives at least one of the logarithm equation problems.

Example: $\log(x)$ has problems when $x \le 0$ and so the second solution is a problem.

So the $x$ where the equations intersect is at $x=\frac{1+\sqrt{5}}{2}$ .

Let's find the $y$ -coordinate.

You may use either equation.

I choose $y=\log(x)$ .

$y=\log(\frac{1+\sqrt{5}}{2})$

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

6 0

2 years ago

Y to the 3rd power times y to the negative 5th power

lina2011 [118]

Y^3 * y^5 = y^8. you add the exponents

8 0

2 years ago