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Yuliya22 [10]
3 years ago
15

The expression to find the perimeter of a rectangle is 2(length + width). What is the perimeter, in units, of a rectangle of len

gth fraction 1 over 3 unit and width fraction 1 over 2 unit?
fraction 2 over 5 unit
fraction 4 over 5 unit
fraction 1 and 1 over 3 units
fraction 1 and 2 over 3 units
Mathematics
2 answers:
likoan [24]3 years ago
7 0

Answer:

The answer is D.) Fraction 1 2/3.

I recently did this test on FLVS

Hope this helped!

Have a wonderful Wednesday!

~Lola

Allisa [31]3 years ago
4 0
First you plug in the data to the formula. So it will be 2(1/3+1/2). You then want to add the fractions together but must first find the common denominator which will be 6. So it is 2(2/6+3/6). The fractions then add up to be 5/6. So all you have to do is multiply it by 2 to get 5/3 which simplifies to 1 2/3 so it is D. 
You might be interested in
Model the equation a over 8 plus 1 equals 4
faltersainse [42]
A/8 +1= 4 because a is devided by 8 then you have to add one to equal 4
8 0
3 years ago
A spherical balloon currently has a radius of 19cm. If the radius is still growing at a rate of 5cm or minute, at what rate is a
miv72 [106K]

Answer:

22670.8 cm³/min

Step-by-step explanation:

Given:

Radius of the balloon at a certain time (r) = 19 cm

Rate of growth of radius is, \frac{dr}{dt}=5\ cm/min

The rate at which the air is pumped in the balloon can be calculated by finding the rate of increase in the volume of the balloon.

So, first we find the volume of the sphere in terms of 'r'. As the balloon is spherical in shape, the volume of the balloon is equal to the volume of a sphere. Therefore,

Volume of balloon is given as:

V=\frac{4}{3}\pi r^3

Now, rate of increase of volume is obtained by differentiating both sides of the equation with respect to time 't'.

Differentiating both sides with respect to time 't', we get:

\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)\\\\\frac{dV}{dt}=\frac{4\pi}{3}(3r^2)(\frac{dr}{dt})\\\\\frac{dV}{dt}=4\pi r^2(\frac{dr}{dt})

Now, plug in 19 cm for 'r', 5 cm per minute for \frac{dr}{dt} and solve for \frac{dV}{dt}. This gives,

\frac{dV}{dt}=4\pi (19 cm)^2(5\ cm/min)\\\\\frac{dV}{dt}=4\times 3.14\times 361\times 5\ cm^3/min\\\\\frac{dV}{dt}=22670.8\ cm^3/min

Therefore, the rate at which the air is being pumped into the balloon is 22670.8 cm³/min.

4 0
3 years ago
8 cm
miss Akunina [59]

Answer:

May be b is the answer...

5 0
2 years ago
Read 2 more answers
Please help I have 9 min
Harrizon [31]
I believe the answer is D because the pathagreom theorum proves it true (excuse my terrible grammar)
5 0
3 years ago
Given: ∆ABC, m∠C = 90° CB = 8, m∠B = 38º Find the area of a circumscribed circle. Find the area of the inscribed circle.
vitfil [10]

Answer:

Circumscribed circle: Around 80.95

Inscribed circle: Around 3.298

Step-by-step explanation:

Since C is a right angle, when the circle is circumscribed it will be an inscribed angle with a corresponding arc length of 2*90=180 degrees. This means that AB is the diameter of the circle. Since the cosine of an angle in a right triangle is equivalent to the length of the adjacent side divided by the length of the hypotenuse:

\cos 38= \dfrac{8}{AB} \\\\\\AB=\dfrac{8}{\cos 38}\approx 10.152

To find the area of the circumscribed circle:

r=\dfrac{AB}{2}\approx 5.076 \\\\\\A=\pi r^2\approx 80.95

To find the area of the inscribed circle, you need the length of AC, which you can find with the Pythagorean Theorem:

AC=\sqrt{10.152^2-8^2}\approx 6.25

The area of the triangle is:

A=\dfrac{bh}{2}=\dfrac{8\cdot 6.25}{2}=25

The semiperimeter of the triangle is:

\dfrac{10.152+6.25+8}{2}\approx 24.4

The radius of the circle is therefore \dfrac{25}{24.4}\approx 1.025

The area of the inscribed circle then is \pi\cdot (1.025)^2\approx 3.298.

Hope this helps!

6 0
3 years ago
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