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3 years ago

Use the percent bar model to find the missing percent

Mathematics

2 answers:

nikdorinn [45]3 years ago

6 0

Answer:

20%

Step-by-step explanation:

In percent bar method, there are two things to be considered.

Part and the whole.

part is considered as the one of the total number of divisions, whereas, the whole is the summation of all the parts.

Now, in the figure, where the missing percent is mentioned, is the first part,

So,

Part = 1

Total = 5

Now as bar represent 100%.

Since, there are 5 partitions, We divide 100 with 5 to determine amount of percent represent by each bar or partition.

Each bar = 100/5 =20%

Since we have to calculate the first bar,

So First bar = 20%,

Note that if we have to calculate two bars, it will be 20 +20 = 40%

But since we have to calculate only first partition so answer is 20%

Phantasy [73]3 years ago

5 0

The answer is 80%
1/5 of 100 is 20.
You find this by dividing 100 by 5 to get 20, then you subtract 20 from 100.

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How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

1 year ago

Find k if (x-3) is a factor of <img src="https://tex.z-dn.net/?f=%20k%5E%7B2%7D%20%20x%5E%7B2%7D%20-kx-2" id="TexFormula1" title

cluponka [151]

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4 0

3 years ago

Read 2 more answers

I thought this was much more easier for you! :)

monitta

Problem 1)

The base of the exponential is 12 which is also the base of the log as well. The only answer choice that has this is choice B.

======================================================================
Problem 2)

log(x) + log(y) - 2log(z)
log(x) + log(y) - log(z^2)
log(x*y) - log(z^2)
log[(x*y)/(z^2)]

Answer is choice D

======================================================================
Problem 3)

log[21/(x^2)]
log(21) - log(x^2)
log(21) - 2*log(x)

This matches with choice B

======================================================================
Problem 4)

Ln(63) = Ln(z) + Ln(7)
Ln(63)-Ln(7) = Ln(z)
Ln(63/7) = Ln(z)
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z = 9

======================================================================
Problem 5)

Ln(5x-3) = 2
5x-3 = e^2
5x = e^2+3
x = (e^2+3)/5

This means choice A is the answer

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3 years ago

Which of the following is the equation for the line perpendicular to the line y=−x+20 that passes through the point (−4, 2)?

coldgirl [10]

I have a feeling it’s a but I’m not that sure good luck might want to wait on other peoples opinions

7 0

3 years ago

Will give branliest and 10 points please answer with correct work

insens350 [35]

Answer: Part A the answer is yes and Part B the answer is no.

Step-by-step explanation:

Part A when you divide 7 by 2 you get 3.5 and if you multiply that number by the rest of the numbers and you get the number beneath the other number.

Part B when you divide 8 by 5 you get 1.6 and if you multiply that number by the rest of the numbers and then you don't get the number beneath the other number.

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3 years ago

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