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2 years ago

7

Difference between

ormula1" title=" \frac{4}{5} - \frac{3}{10} " alt=" \frac{4}{5} - \frac{3}{10} " align="absmiddle" class="latex-formula">

2 answers:

pashok25 [27]2 years ago

5 0

$\\ \sf\longmapsto \dfrac{4}{5}-\dfrac{3}{10}$

Take LCM

$\\ \sf\longmapsto \dfrac{8-3}{10}$

$\\ \sf\longmapsto \dfrac{5}{10}$

Cancel by 5

$\\ \sf\longmapsto \dfrac{1}{2}$

vampirchik [111]2 years ago

3 0

Step-by-step explanation:

see below

$\longrightarrow \: \frac{4}{5} - \frac{3}{10}$

$\longrightarrow \: \frac{8 - 3}{10}$

$\longrightarrow \: \cancel{\frac{5}{10} }$

$\longrightarrow \: \frac{1}{2}$

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Camilla's employer pays 90% of the total cost of her family's health insurance

Irina-Kira [14]

Answer:

B. $10

Step-by-step explanation:

Camila's employer pays 90% of the total cost of her family's health insurance

plan

Therefore, Camilla pays only 10%

If the total cost of the plan increases from $850 per month to $950 per month, how much more will Camilla have to pay

each month?

Amount of health insurance Camilla pays at $850 per month = 10% of $850

= 0.1 * $850

= $85

Amount of health insurance Camilla pays at $950 per month = 10% of $950

= 0.1 * $950

= $95

Amount Camilla pays more = $95 - $85

= $10

B. $10

8 0

3 years ago

Pretest: Unit 1

Alex_Xolod [135]

A vertical line that the graph of a function approaches but never intersects. The correct option is B.

<h3>When do we get vertical asymptote for a function?</h3>

Suppose that we have the function f(x) such that it is continuous for all input values < a or > a and have got the values of f(x) going to infinity or -ve infinity (from either side of x = a) as x goes near a, and is not defined at x = a, then at that point, there can be constructed a vertical line x = a and it will be called as vertical asymptote for f(x) at x = a

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3 0

1 year ago

What is the ratio of the area of the inner square to the area of the outer square?

Anastasy [175]

Answer:

$\frac{(a-b)^2+b^2}{a^2}$

Step-by-step explanation:

Since, By the given diagram,

The side of the inner square = Distance between the points (0,b) and (a-b,0)

$=\sqrt{(a-b-0)^2+(0-b)^2}$

$=\sqrt{(a-b)^2+b^2}$

Thus the area of the inner square = (side)²

$=(\sqrt{(a-b)^2+b^2})^2$

$=(a-b)^2+b^2\text{ square cm}$

Now, the side of the outer square = Distance between the points (0,0) and (a,0),

$=\sqrt{(a-0)^2+0^2}$

$=\sqrt{a^2}=a$

Thus, the area of the outer square = (side)²

$=a^2\text{ square cm}$

Hence, the ratio of the area of the inner square to the area of the outer square

$=\frac{(a-b)^2+b^2}{a^2}$

4 0

3 years ago

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Anna35 [415]

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2 years ago

Kevin had a total of $320 to spend while shopping. He had spent $156 on video games. He wants to buy caps that are on sale for $

zloy xaker [14]

Maximum number of caps that Kevin can buy is 10.

Solution:

Total money for shopping = $320

Amount spent for video game = $156

Remaining amount that Kevin had = $320 – $156

= $164

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$\text{Number of caps can buy }=\frac{\text { Remaining amount that Kevin had }}{\text { cost of each cap }}$

$$=\frac{164}{15}$

$$=10\frac{14}{15}$

Maximum number of caps that Kevin can buy is 10.

8 0

3 years ago