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3 years ago

What is the area of the square below ?

Mathematics

1 answer:

ivolga24 [154]3 years ago

6 0

To find the are of a square, we simply square the value of a side
so..
${5}^{2} = area$
Area = 25 square units

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Given that f(x)=2x/3 evaluate f^-1(6)

abruzzese [7]

Answer:

Step-by-step explanation:

we know by definition that

$f^{-1}(f(x))=x\\\\f^{-1}(\frac{2x}{3})=x\\\\f^{-1}(x)=ax\\\\f^{-1}(\frac{2x}{3})=a(\frac{2x}{3})=x\\\\f^{-1}(x)=\frac{3x}{2}$

so now we evaluate

$f^{-1}(6)=\frac{3*6}{2}\\\\=3*3\\\\f^{-1}(6)=9$

and if we want to do an extra step

$f^{-1}(f(6))=6\\\\f^{-1}(\frac{2*6}{3})=f^{-1}(4)=\frac{3*4}{2}=6$

which works.

4 0

2 years ago

The price of an item has risen to $91 today. yesterday it was $65 l. find the percentage increase

Natali [406]

Answer:

40%

Step-by-step explanation:

First you take $19 - $65 to get $26

Then: 26 ×100= 2600÷ 65= 40%

8 0

3 years ago

Hawick is 1515 miles south of Abbotsford, and Kelso is 1717 miles east of Abbotsford. What is the distance from Hawick to Kelso?

Mashcka [7]

Answer:

As per the statement:

Hawick is 15 miles south of Abbotsford, and Kelso is 17 miles east of Abbotsford.

Let H represents Hawick , A represents Abbotsford and K represents Kelso

See the diagram as shown below:

Distance of AH = 15 miles

Distance of AK = 17 miles.

We have to find the distance HK:

Using Pythagoras theorem;

$\text{Hypotenuse side}^2 = \text{Adjacent side}^2 + \text{opposite side}^2$

then;

$\text{HK}^2 = 15^2+17^2 = 225 + 289 = 514$

$HK = \sqrt{514} =22.671581$ miles.

Therefore, the distance from Hawick to Kelso( to the nearest tenth place) is 22.6 miles

5 0

3 years ago

Read 2 more answers

Could someone answer and explain these please? Thank you!

Oksana_A [137]

Answer 1:

It is given that the positive 2 digit number is 'x' with tens digit 't' and units digit 'u'.

So the two digit number x is expressed as,

$x=(10 \times t)+(1 \times u)$

$x=10t+u$

The two digit number 'y' is obtained by reversing the digits of x.

So, $y=(10 \times u)+(1 \times t)$

$y=10u+t$

Now, the value of x-y is expressed as:

$x-y=(10t+u)-(10u+t)$

$x-y=10t+u-10u-t$

$x-y=9t-9u$

$x-y=9(t-u)$

So, $9(t-u)$ is equivalent to (x-y).

Answer 2:

It is given that the sum of infinite geometric series with first term 'a' and common ratio r<1 = $\frac{a}{1-r}$

Since, the sum of the given infinite geometric series = 200

Therefore, $\frac{a}{1-r}=200$

Since, r=0.15 (given)

$\frac{a}{1-0.15}=200$

$\frac{a}{0.85}=200$

$a=0.85 \times 200$

a=170

The nth term of geometric series is given by $ar^{n-1}$ .

So, second term of the series = $ar^{2-1}$ = ar

Second term = $170 \times 0.15$

= 25.5

So, the second term of the geometric series is 25.5

Step-by-step explanation:

8 0

3 years ago

This is the range of sample statistics obtained when many samples are selected?

Leno4ka [110]

The range of sample statistics obtained when many samples are selected is a sampling distribution. A sampling distribution is also a probability distribution of the statistics obtained in the large number of samples drawn from a specific population. It also shows every possible result that the statistic can take in every possible sample from the population and how often will each results happens.

4 0

3 years ago