All categories

2 years ago

Hi does anyone know this it is important please help me due tonight i will do anything

Mathematics

2 answers:

VikaD [51]2 years ago

5 0

Answer:

Hewo Asuna here

Answer is $20

Step-by-step explanation:

Hope this helps

Inessa [10]2 years ago

4 0

Answer:

$20

Step-by-step explanation:

Total cost of items bought

= $31.21 +$12.75 +$13.25

= $57.21

Remaining cost after paying $40

= $57.21 -$40

= $17.21

Amount borrowed

= change +$17.21

= $2.79 +$17.21

= $20

You might be interested in

In a right triangle, a and b are the lengths of the legs and c is the length of the hypotenuse. If a=8.9 meters and b=2.9 meters

Temka [501]

The answer is 9 to the nearest tenth

5 0

2 years ago

If the measure of a central angle is 213 degrees, then the measure of the arc it intercepts is

Leto [7]

Answer:

$\Delta s = \frac{71\pi}{60}\cdot r$

Step-by-step explanation:

Let r be the radius of the circle, the measure of the arc is equal to:

$\Delta s = \frac{213^{\circ}}{360^{\circ}}\cdot 2\pi \cdot r$

$\Delta s = \frac{71\pi}{60}\cdot r$

8 0

3 years ago

How much string would you need to go around the bottom of a hat shaped like a cone if the radius is 4 inches? (Use 3.14 for π)

Softa [21]

Answer:

25.12 inches

Step-by-step explanation:

To find the circumference or the perimeter of a circle it is the diameter times pi. (pi = 3.14) Or you can do radius times 2 times pi.

Answer is 4 times 2 times 3.14 which is 25.12

7 0

3 years ago

Problem Solving Skills

____ [38]

Answer:i think the answer is A(x/.ly/

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

PLEASE HURRY IM TIMED!!!

andriy [413]

Answer-

A. Brandon’s sound intensity level is 1/4th as compared to Ahmad’s.

Solution-

Given that, loudness measured in dB is

$L=10\log \frac{I}{I_0}$

Where,

I = Sound intensity,

I₀ = 10⁻¹² and is the least intense sound a human ear can hear

Given in the question,

I₁ = Intensity at Brandon's = 10⁻¹⁰

I₂ = Intensity at Ahmad's = 10⁻⁴

Then,

$L_1=10\log \frac{I_1}{I_0}=10\log \frac{10^{-10}}{10^{-12}}=10\log \frac{1}{10^{-2}}=10\log 10^{2}=2\times10\log 10=20$

$L_2=10\log \frac{I_2}{I_0}=10\log \frac{10^{-4}}{10^{-12}}=10\log \frac{1}{10^{-8}}=10\log 10^{8}=8\times10\log 10=80$

$\therefore \frac{L_1}{L_2} =\frac{20}{80} =\frac{1}{4}$

5 0

3 years ago