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3 years ago

1. 2/4 x 2/9 2. 3/4 x 4/9 3. 2/5 x 5/8

Mathematics

1 answer:

tamaranim1 [39]3 years ago

6 0

1. 1/9
2. 1/3
3. 1/4
ensure when you multiply fractions to simplify the fraction to its lowest terms!!

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The ratio of areas between two similar triangles is 1:4. if one side of the smaller triangle is 2 units, find the measure of the

dimaraw [331]

1:4
smaller side = 2
bigger side = 8 since 2/8 = 1/4
1/4 = 2/x
x=4*2=8

5 0

3 years ago

I am so stuck! :( Pls help me!!! I don't get the problem and I don't know how to solve it

Lelu [443]

Answer:

$\frac{n^{2}}{2a}$

Step-by-step explanation:

Just split it up first $\frac{9an^{3}}{18a^{2}n}=\frac{9}{18}\times \frac{a}{a^2} \times\frac{n^3}{n}$
Then 9 divided by 18 can be simplified into the fraction 1/2
What can go into $a$ and $a^2$ ? a itself right, so if you divide a by a you get 1, if you divide $a^2$ by a you get a
Same as the a fraction, if you divide $n^3$ by n you get $n^2$
$\frac{9an^{3}}{18a^{2}n}=\frac{9}{18}\times \frac{a}{a^2} \times\frac{n^3}{n}=\frac{1}{2}\times \frac{1}{a} \times\frac{n^2}{1}=\frac{n^{2}}{2a}$

4 0

3 years ago

Read 2 more answers

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10

zvonat [6]

The approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

$FV=P\left(1-\dfrac{r}{100}\right)^n$

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is $n_1$ . Thus, by the above formula for the first car,

$0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}$

Take log both the sides as,

$\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85$

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is $n_2$ . Thus, by the above formula for the second car,

$0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}$

Take log both the sides as,

$\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14$

The difference in the ages of the two cars is,

$d=4.85-3.14\\d=1.71\rm years$

Thus, the approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0

2 years ago

Write an equation of the line that passes through (2,−5) and is parallel to the line 2y=3x+10

LenaWriter [7]

Step-by-step explanation:

Divide two on both sides to get rid of it and the make the equation in the form y = mx + c

$\frac{2}{2} y = \frac{3}{2} x + \frac{10}{2}$

y = 3/2x + 5

since both lines are parallel they must have the same gradient which is 3/2

y = 3/2x + c

All you have to do now is to replace x and y with (2, -5) to find c

x = 2

y = -5

-5 = 3/2 × 2 + c

-5 = 3 + c

c = -5-3

c = -8

$y = \frac{3}{2} x - 8$

8 0

3 years ago

Find the x and y intercepts. 3x - 2y = 20

Alex787 [66]

Solutions: x=1<span>. Hope its right :)</span>

6 0

4 years ago