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Amiraneli [1.4K]

3 years ago

8

When (8668/25) + (4141/9) - (5533/25) is computed and written as a mixed number in simplest form, what is the fractional part of

that mixed number?

1 answer:

podryga [215]3 years ago

8 0

This is the answer when written in mixed form if that's what the question meant.
$585 \ \frac{23}{45}$

You might be interested in

Which table represents exponential growth? A 2-column table has 4 rows. The first column is labeled x with entries 1, 2, 3, 4. T

Papessa [141]

Answer:

Table 2

Step-by-step explanation:

We have the tables:

<u>Table 1:</u>

x: 1 2 3 4

y: 2 4 6 8

<u>Table 2:</u>

x: 1 2 3 4

y: 2 4 8 16

<u>Table 3:</u>

x: 1 2 3 4

y: 2 4 7 11

<u>Table 4:</u>

x: 1 2 3 4

y: 2 4 6 10

An exponential growth data set will show a common ratio between y values. Let's look at each of the ratios from each table.

<u>Table 1:</u>

8/6 = 4/3

6/4 = 3/2

Already, we can see that 4/3 ≠ 3/2, which means that this doesn't have a common ratio. So Table 1 is wrong.

<u>Table 2:</u>

16/8 = 2

8/4 = 2

4/2 = 2

The common ratio here is 2, so we know this is correct.

<u>Table 3:</u>

11/7 = 1.57

7/4 = 1.75

Again, we can see that 1/57 ≠ 1.75, so this is wrong.

<u>Table 4:</u>

10/6 = 1.67

6/4 = 1.5

Again, there is no common ratio here, so this is wrong.

The answer is thus Table 2.

8 0

3 years ago

Read 2 more answers

HDTV screens usually have a width-to-heigh ratio of 16:9. Which screen dimensions are in the same proportion?

Lana71 [14]

Answer:

4: 3

Step-by-step explanation:

In this case, we have that there is a relationship between width and height, which is 16: 9, which means that any screen that has the same ratio would be an identical screen.

How would we do this? it must either be raised to a power or rationalized on both sides of the ratio.

If we raise to a power, it would give very large dimensions for an HDTV, therefore, it is best to remove the square root on both sides, since it has more common values for a TV.

$\sqrt{16} :\sqrt{9}$

4: 3

Therefore, a 4: 3 screen is a good choice for an HDTV with the same ratio.

5 0

2 years ago

Otis plans to make 3 batches of sidewalk chalk How much plaster of paris does he need

den301095 [7]

Answer:

$4\frac{1}{2}$ cups plaster of paris.

Step-by-step explanation:

Given:

for one batch of sidewalk chalk = $1\frac{1}{2}$ cup plaster of paris.

Number of bathes = 3

We need to find quantity of plaster of paris.

Solution:

For making 1 batch of sidewalk chalk = $1\frac{1}{2}$ cup plaster of paris.

For making 3 batches of sidewalk chalk = $3\times 1\frac{1}{2}$ cup plaster of paris.

$= 3\times \frac{2\times 1+1}{2}$

$= 3\times \frac{2+1}{2}$

$= 3\times \frac{3}{2}$

$= \frac{3\times 3}{2}$

$= \frac{9}{2}$

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

Therefore, he needed $4\frac{1}{2}$ cups plaster of paris.

5 0

3 years ago

You are riding a roller coaster when a shoe falls off your foot. The function y=300−16t2 represents the height y (in feet) of th

Juli2301 [7.4K]

Answer:

Time to land on 11ft is 4.25 seconds

Step-by-step explanation:

Given

$y = 300 - 16t^2$

Required

Solve for t when height is 11ft

This implies that $y = 11$

So: substitute 11 for y in $y = 300 - 16t^2$

$11 = 300 - 16t^2$

Collect Like Terms

$16t^2 = 300 - 11$

$16t^2 = 289$

Express both sides as squares

$4^2 * t^2 = 17^2$

$(4t)^2 = 17^2$

Take square roots of both sides (ignore negative)

$4t = 17$

Solve for t

$t = \frac{17}{4}$

$t = 4.25$

Hence, time to land on 11ft is 4.25 seconds

8 0

2 years ago

Look at the graph shown below:

ElenaW [278]

So.. take a peek at the picture... let's get two points from it, hmm say 0,4 notice it touches the y-axis there, and say hmmm -4, 1, almost at the bottom of the line

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 0}}\quad ,&{{ 4}})\quad 
% (c,d)
&({{ -4}}\quad ,&{{ 1}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-4}{-4-0}$

$\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad 
\begin{array}{llll}
\textit{plug in the values for }
\begin{cases}
y_1=4\\
x_1=0\\
m=\boxed{?}
\end{cases}\\
\textit{and solve for "y"}
\end{array}\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}$

once you get the slope and solve for "y", that'd be the equation of the line.

7 0

2 years ago