What is 1/5 reduce to its lowest term

Answers

Answer 1

Answer: 1/5 is the lowest term because the numerator has 1

Answer 2

Answer: 1/5 reduced is just 1/5 because if anything has a 1 in it's numerator you know that the GCF is just 1 so you can't have a even more simplified form of this.

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For the function y = 4x – 11, what is the output when the input is 10?

Answers

= 4x - 11

x is the input, y is the output.

So if you want to know what input is needed to get an output of 10, set y to 10 and solve for x:

10 = 4x - 11

21 = 4x

x = 21/4

Find the volume and the lateral area of a frustum of a right circular cone whose radii are 4 and 8 cm, and slant height is 6 cm.A chimney, 100 ft. high, is in the form of a frustum of a right circular cone with radii 4 ft. and 5 ft. Find the lateral surface area of the chimney.
The volume of a frustum of a right circular cone is 52π ft3. Its altitude is 3 ft. and the measure of its lower radius is three times the measure of its upper radius. Find the lateral area of the frustum.
A frustum of a right circular cone has an altitude of 24 in. If its upper and lower radii are 15 in. and 33 in., respectively, find the lateral area and volume of the frustum.
In a frustum of a right circular cone, the radius of the upper base is 5 cm and the altitude is 8√3cm. If its slant height makes an angle of 60° with the lower base, find the total surface area of the frustum.
A water tank in the form of an inverted frustum of a cone has an altitude of 8 ft., and upper and lower radii of 6 ft. and 4 ft., respectively. Find the volume of the water tank and the wetted part of the tank if the depth of the water is 5 ft.
The total surface area of a frustum of a right circular cone is 435π cm2, and the base areas are 81π cm2 and 144π cm2. Find the slant height and the altitude of the frustum.
The base edges of a frustum of a regular pentagonal pyramid are 4 in. and 8 in., and its altitude is 10 in. Find the volume and the total area of the frustum.
Find the volume of a frustum of a regular square pyramid if the base edges are 14 cm and 38 cm, and the measure of one of its lateral edges is 24 cm.
Find the volume of a frustum of a regular square pyramid if the base edges are 7 cm and 19 cm, and the lateral edge is inclined at an angle of 60° with the lower base.
Find the volume of a frustum of a regular square pyramid if the base edges are 13 cm and 29 cm, and the lateral edge is inclined at an angle of 45° with the lower base.
The base edges of a frustum of a regular square pyramid measure 20 cm and 60 cm. If one of the lateral edges is 75 cm, find the total surface area of the frustum.
A frustum of a regular hexagonal pyramid has an upper base edge of 16 ft. and a lower base edge of 28 ft. If the lateral area of the frustum is 1,716 ft.2, find the altitude of the frustum.
A regular hexagonal pyramid has an upper base edge of 16 ft. and a lower base edge 28 ft. If the volume of the frustum is 18,041 ft.3, find the lateral area of the frustum.
The lateral area of a frustum of a regular triangular pyramid is 1,081 cm2, and the altitude and lateral edge are 24 cm and 26 cm, respectively. Find the lengths of the sides of the bases.

Answers

the complete answers in the attached figure

Part 1) we have

$r=4cm\n R=8 cm\n L=6cm$

Find the height h

$h^(2)=L^(2) -(R -r)^(2)\n h^(2)=6^(2) -(8-4)^(2)\n h^(2)=36-16\n h=√(20) cm$

Find the volume

$V=(1)/(3)\pi[R^(2) +r^(2) +Rr]h\n\n V=(1)/(3)\pi[8^(2) +4^(2) +8*4]√(20)\n \n V=(1)/(3)\pi[112]√(20)\n \n V=524.52 cm^(3)$

Find the lateral area

$LA=\pi (R+r)L\n LA=\pi *(8+4)*6\n LA=226.19 cm^(2)$

the answer Part 1) is

a) the volume is equal to $524.52 cm^(3)$

b) The Lateral area is equal to $226.19 cm^(2)$

Part 2) we have

$r=4ft\n R=5 ft\n h=100 ft$

Find the slant height L

$L^(2)=h^(2)+(R -r)^(2)\n L^(2)=100^(2) +(5-4)^(2)\n L^(2)=10000+1\n L=√(10001) ft$

Find the lateral area

$LA=\pi (R+r)L\n LA=\pi *(5+4)*√(10001)\n LA=2,827.57 ft^(2)$

the answer part 2) is

a) The Lateral area is equal to $2,827.57 ft^(2)$

Part 3) we have

$V=52\pi ft^(3) \n h=3ft\n R=3r$

Step 1

Find the values of R and r

$V=(1)/(3)\pi[R^(2) +r^(2) +Rr]h$

substitute $R=3r$ in the formula above

$V=(1)/(3)\pi[(3r)^(2) +r^(2) +(3r)*r]*3$

$V=(1)/(3)\pi[7r)^(2)]*3$

$V=[tex] 52\pi$

$52\pi =\pi [7r^(2) ]\n r^(2) =(52)/(7) \n \n r=2.73 ft$

$R=3*2.73\n R=8.19 ft$

Step 2

Find the slant height L

$L^(2)=h^(2)+(R -r)^(2)\n L^(2)=3^(2) +(8.19-2.73)^(2)\n L^(2)=38.81\n L=6.23 ft$

Step 3

Find the lateral area

$LA=\pi (R+r)L\n LA=\pi *(8.19+2.73)*6.23 LA=213.73 ft^(2)$

the answer Part 3) is

a) The lateral area is equal to $213.73 ft^(2)$

Part 4) we have

$r=15 in\n R=33 in\n h=24 in$

Find the slant height L

$L^(2)=h^(2)+(R -r)^(2)\n L^(2)=24^(2) +(33-15)^(2)\n L^(2)=576+324\n L=30 in$

Find the lateral area

$LA=\pi (R+r)L\n LA=\pi *(33+15)*30\n LA=4,523.89 in^(2)$

Find the volume

$V=(1)/(3)\pi[R^(2) +r^(2) +Rr]h\n\n V=(1)/(3)\pi[33^(2) +15^(2) +33*15]24\n \n V=(1)/(3)\pi[112]24\n \n V=142.83 in^(3)$

the answer is

a) The lateral area is equal to $4,523.89 in^(2)$

b) the volume is equal to $142.83 in^(3)$

Part 5) we have

$r=5 cm\n h=8√3 cm$

Step 1

Find the value of (R-r)

$tan 60=√(3)$

$tan 60=((R-r))/(8√(3)) \n\n R-r= √(3) *8√(3) \n R-r=24 cm\n R=24+r\n R=24+5\n R=29 cm$

Step 2

Find the value of slant height L

$L^(2)=h^(2)+(R -r)^(2)\n L^(2)=(8√(3))^(2)+(24-5)^(2)\n L^(2)=192+361\n L=23.52 cm$

Step 3

Find the lateral area

$LA=\pi (R+r)L\n LA=\pi *(24+5)*23.52\n LA=2,142.82 cm^(2)$

Step 4

Find the total area

total area=lateral area+area of the top+area of the bottom

Area of the top

$r=5 cm\n A=\pi *r^(2) \n A=\pi *25\n A=78.54 cm^(2)$

Area of the bottom

$r=24 cm\n A=\pi *r^(2) \n A=\pi *576\n A=1,809.56 cm^(2)$

Total surface area

$SA=2,142.82+78.54+1,809.56\n SA=4,030.92 cm^(2)$

the answer is

a) The total surface area is $4,030.92 cm^(2)$

Part 6)

Part a) Find the volume of the water tank

we have

$r=4 ft\n R=6 ft\n h=8 ft$

Step 1

Find the volume

$V=(1)/(3)\pi[R^(2) +r^(2) +Rr]h\n\n V=(1)/(3)\pi[6^(2) +4^(2) +6*4]8\n \n V=(1)/(3)\pi[76]8\n \n V=636.70 ft^(3)$

the answer Part a) is $636.70 ft^(3)$

Part b) Find the volume of the wetted part of the tank if the depth of the water is 5 ft

by proportion find the radius R of the upper side for h=5 ft

$((R1-r))/(8) =((R2-r))/(5) \n\n ((6-4))/(8) =((R2-4))/(5)\n \n(R2-4)= 1.25\n R2=4+1.25\n R2=5.25 ft$

Find the volume for $R2=5.25 ft$

$V=(1)/(3)\pi[R^(2) +r^(2) +Rr]h\n\n V=(1)/(3)\pi[5.25^(2) +4^(2) +5.25*4]5\n \n V=(1)/(3)\pi[64.56]5\n \n V=338.05 ft^(3)$

the answer Part b) is $338.05 ft^(3)$

Part 7) we have

$SA=435\pi cm^(2) \n A1=144\pi cm^(2)\n A2=81\pi cm^(2)$

Step 1

Find the value of R and the value of r

$A1=\pi *R^(2) \n 144\pi =\pi *R^(2)\n R=12 cm$

$A2=\pi *r^(2) \n 81\pi =\pi *r^(2)\n r=9 cm$

Step 2

Find the value of lateral area

$LA=SA-A1-A2\n LA=435\pi -144\pi -81\pi \n LA=210\pi cm^(2)$

Step 3

Find the slant height

$LA=\pi (R+r)L\n\n L=(LA)/(\pi(R+r)) \n \n L=(210\pi)/(\pi(12+9)) \n \n L=10 cm$

Find the altitude of the frustum

$h^(2) =L^(2) -(R-r)^(2) \n h^(2) =10^(2) -(12-9)^(2)\n h^(2)=91\n h=9.54 cm$

the answer Part a) is

the slant height is $10 cm$

the answer Part b) is

the altitude of the frustum is $9.54 cm$

Find the volume and the lateral area of a frustum of a right circular cone whose radii are 4 and 8 cm, and slant height is 6 cm.
$h= √(s^2-(R_1-R_2)^2) \n = √(6^2-(4-8)^2) \n = √(36-16) \n = √(20)$
$Volume= (1)/(3) \pi h(R_1^2+R_1R_2+R_2^2) \n = (1)/(3) \pi * √(20) (4^2+4 * 8+8^2) \n = (1)/(3) \pi √(20) (16+32+64) \n = (1)/(3) \pi √(20) (112) \n =524.5cm^3$
Lateral area = Total surface area - area of base - area of top
$Lateral \ area= \pi (R_1+R_2)s \n = \pi (4+8) * 6 \n =12 \pi * 6 \n =72 \pi \n =226.2cm^2$

A cube has a surface area of 72 in2.Find the surface area of a scaled image with a scale factor of 1.5.

in2

Answers

so assuming scaled down it is
1:1.5 or
2:3
so
72 is the 3 units
divide 72 by 3
72/3=14
14 times 2=28

the surface area=28 in^2

What is 6.42 written as a fraction in lowest terms?

Answers

Answer:

6 21/50

Step-by-step explanation:

Convert 6.42 into a fraction (6 42/100)
The GCF of 42 and 100 is 2, so divide 42 and 100 by 2 to simplify.

Answer:

6 21

...50

Step-by-step explanation:

Find b, given that a = 18.2, B = 62°, and C = 48°. Round answers to the nearest whole number. Do not use a decimal point or extra spaces in the answer or it will be marked incorrect.

Answers

Answer:

Step-by-step explanation:

We have been given that in triangle ABC, measure of angle B is 62 degrees and measure of angle C is 48 degrees. The length of side opposite to angle a is 18.2. We are asked to find length of side b.

We will use law of sines to solve for side b.

$\frac{a}{\text{sin}(A)}=\frac{b}{\text{sin}(B)}=\frac{c}{\text{sin}(C)}$

$m\angle A+m\angle B+m\angle C=180^(\circ)\n\nm\angle A+62^(\circ)+48^(\circ)=180^(\circ)$

$m\angle A+110^(\circ)=180^(\circ)$

$m\angle A+110^(\circ)-110^(\circ)=180^(\circ)-110^(\circ)$

$m\angle A=70^(\circ)$

Upon substituting our given values, we will get:

$\frac{18.2}{\text{sin}(70^(\circ))}=\frac{b}{\text{sin}(62^(\circ))}$

$\frac{18.2}{\text{sin}(70^(\circ))}*\text{sin}(62^(\circ))=\frac{b}{\text{sin}(62^(\circ))}*\text{sin}(62^(\circ))$

$\frac{18.2}{\text{sin}(70^(\circ))}*\text{sin}(62^(\circ))=b$

$b=\frac{18.2}{\text{sin}(70^(\circ))}*\text{sin}(62^(\circ))$

$b=(18.2)/(0.939692620786)*0.882947592859$

$b=19.1551999*0.882947592859$

$b=16.9130376$

$b\approx 17$

Therefore, the length of side b is 17 units.

A credit card issuer charges an APR of 15.77%, and its billing cycle is 30 days long. What is it periodic interest rate?

Answers

Given:
Billing cycle = 30 days long
APR = 15.77%

There is 365 days in a year.

365 / 30 = 12.17

15.77% / 12.17 = 1.2958% or 1.30%

The periodic interest rate is 1.30%.

Answer:

1.30%

Step-by-step explanation:

I just answered it

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