All categories

2 years ago

Find the area of a circle whose radius is 2.5cm

Mathematics

2 answers:

Marizza181 [45]2 years ago

6 0

Answer:

19.6349540849

Step-by-step explanation:

π × 2.5^2 = 19.6349540849

Marina CMI [18]2 years ago

4 0

2.5 squared is 6.25
pi multiplied by 6.25 is 19.6
The area of a circle whose radius is 19.6cm

You might be interested in

Is 40 a multiple of 8

Natasha2012 [34]

Yes because,
8×1 = 8
8×2 = 16
8×3 = 24
8×4 = 32
8×5 = 40

So 40 is a multiple of 8

4 0

3 years ago

a library has an average of 510 visitors on Sundays and 240 visitors on other days. The average number of visitors per day in a

serious [3.7K]

Answer:

285

Step-by-step explanation:

personally this is how i would do this question but there are other ways.

i would start off by seeing how many sundays there are in the month (which is 5) then the number of other days which is 25. i'd then multiply 5 by 510 which is 2550. add that to 240 times 25 (which is 6000) and all of that together is 8550. divide that by the number of the days in the month (30) and then you'll get the daily avergae for the month

hope this helped:)

6 0

3 years ago

How do I calculate 5 percent of 60 using fractions

irinina [24]

60*5%=60*5/100=300/100=3
So the answer is 3

5 0

3 years ago

Read 2 more answers

Let f(x)equals=33xminus−1, h(x)equals=startfraction 7 over x plus 5 endfraction 7 x+5 . find (hcircle◦f)(66).

Ronch [10]

$\bf \begin{cases}
f(x)=3x-1\\
h(x)=\cfrac{7}{x+5}\\\\
(h\circ f)(x)=h(~~f(x)~~)
\end{cases}
\\\\\\
f(66)=3(66)-1\implies \boxed{f(66)=197}
\\\\\\
\stackrel{h(~~f(x)~~)}{h(~~f(66)~~)}\implies \stackrel{h(~~f(66)~~)}{h\left(~~\boxed{197}~~ \right)}=\cfrac{7}{\boxed{197}+5}\implies h\left(~~\boxed{197}~~ \right)=\cfrac{7}{202}$

7 0

3 years ago

Use the discriminant to determine the number of solutions to the quadratic equation 3x^2+5x=-1

kari74 [83]

Answer:

Two real distinct solutions

Step-by-step explanation:

Hi there!

Background of the Discriminant

The discriminant $b^2-4ac$ applies to quadratic equations when they are organised in standard form: $ax^2+bx+c=0$ .

All quadratic equations can be solved with the quadratic formula: $x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}}$ .

When $b^2-4ac$ is positive, it is possible to take its square root and end up with two real, distinct values of x.

When it is zero, we won't be taking the square root at all and we will end up with two real solutions that are equal, or just one solution.

When it is negative, it is impossible to take the square root and we will end up with two non-real solutions.

Solving the Problem

$3x^2+5x=-1$

We're given the above equation. It hasn't been organised completely in $ax^2+bx+c=0$ , but we can change that by adding 1 to both sides to make the right side equal to 0:

$3x^2+5x+1=0$

Now that we can identify the values of a, b and c, we can plug them into the discriminant:

$D=b^2-4ac\\D=(5)^2-4(3)(1)\\D=25-4(3)(1)\\D=25-12\\D=13$

Therefore, because the discriminant is positive, the equation has two real, distinct solutions.

I hope this helps!

4 0

3 years ago