What is the value of 7 to the fifth power?

Answers

Answer 1

Answer:

Hi there!

Since we know a number raised to a power means exponents, and exponents mean repeated multiplication, the following is what we would do --

7 × 7 × 7 × 7 × 7 = 16,807

So, 7 raised to the 5th power equals 16,807.

Hope this helps!

Message me if you need anything, I'd be happy to help! :D

Answer 2

Answer:

The value of 7 raised to the fifth power ( $7^5$ ) is 16,807.

To calculate 7 raised to the fifth power, you multiply 7 by itself five times.

7⁵

= 7 × 7 × 7 × 7 × 7

Simplifying the calculation:

= 16,807

So, the value of 7 to the fifth power is 16,807.

Learn more about Mathematical operations here:

brainly.com/question/20628271

#SPJ6

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An equation is shown below:2(3x − 5) = 1

Which of the following correctly shows the steps to solve this equation?

Step 1: 6x − 10 = 1; Step 2: 6x = 11
Step 1: 6x − 5 = 1; Step 2: 6x = 6
Step 1: 5x − 3 = 1; Step 2: 5x = 4
Step 1: 5x − 7 = 1; Step 2: 5x = 8

Answers

The option that shows the steps to solve the given equation is A) and this can be determined by using the arithmetic operations.

Given :

Linear Equation -- 2(3x − 5) = 1

The following steps can be used in order to evaluate the given linear equation:

Step 1 - The arithmetic operations can be used in order to evaluate the given linear equation.

Step 2 - Write the given linear equation.

2(3x − 5) = 1

Step 3 - Multiply 2 by (3x - 5) in the above equation.

6x - 10 = 1

Step 4 - Add 10 on both sides in the above expression.

6x - 10 + 10 = 1 + 10

6x = 11

Step 5 - Divide both sides by 6 in the above equation.

x = 11/6

From the above steps, it can be concluded that the correct option is A).

For more information, refer to the link given below:

brainly.com/question/25834626

i would have to go with the first one because 2(3x-5)=1 is the same as 6x-10=1
then all you have to do is divide 11 by 6 and the answetr should be c or in this case x=1.83
to check your work do this 6(1.83)-10 and your answer should come to .98
all you have to do is round to 1.0

an apple orchard sells apples in a bags of 10. the orchard sold a total of 20,430 apples in one day. how many bags of apples was it

Answers

Answer:

2043

Step-by-step explanation:

you divide 20430 by 10 and you get your answer

Find 7% of 280
Find 12% of 300
Find 2% of $1250

Answers

percent means parts out of 100
7%=7/100=0.07
'of' means multiply so

7%=0.7
7% of 280=0.07 times 280=196

12%=0.12
12% of 300=36

2%=0.02
2% of $1250=$25

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.
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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
$Lateral \ area= \pi (R_1+R_2)s \n = \pi (4+8) * 6 \n =12 \pi * 6 \n =72 \pi \n =226.2cm^2$

Work out -4x-3+2x-6 can I have help please ?

Answers

Let's reorder the equation. -4x+2x-3-6. Now we can reduce like terms. -4x+2x= -2x and -3-6= -9. Put the two back together and it is simplified to -2x-9.

-4x-3+2x-6-4x+2x-6-3-2x-9I hope that is good!

A ratio of men to women in a party is exactly 3;2 if there are a total of 120 people at the party how many of them are women ? Can someone please explain to me how the answer is 48

Answers

If the ratio is 3:2 then you add that together to get 5.Which tell you what one group would equal. Then you divide 120 by 5 to get an answer of 24. If women is equal to 2 out of a total of 5, then you multiply 24 by 2 to get 48.

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