Please help brainiest will give you!Find the slope of it!
A. -3
B. -1/3
C. 1/3
D. 3

Answers

Answer 1

Answer: The correct answer is B

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Manny makes dinner using 1 box of pasta and 1 jar of sauce. if pasta is sold in packages of 6 boxes and sauce is sold in packages of 3 jars, what is the least number of dinners that Manny can make without any supplies leftover?

Answers

You need to find the least common multiple of 6 and 3. This the minimum whole number that contains the number of boxes in one package of pasta and the number of jars in one package of sauce. t he least common multiple of 6 and 3 is 6. This means that Manny can make 6 dinners, which will be made with 1 package of pasta (6 boxes ) and two packages of sauce (2 * 3 = 6 jars).

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

Simplify (3x + 5) + (2x - 9) - (4x + 3).

Answers

The answer is x-7
(3x+2x-4x)+(5+-9-3)
(5x-4x)+(-4-3)
(1x)+(-7)
x-7

Function f(x) is represented on the graph. Which statement identifies the effect of replacing f(x) with 1/2f(x) on the graph? A)The curve would remain the same size but would be flipped upside down. B)The curve would remain the same size but would shift to the left. C)The curve would be narrower, but the vertex would be in the same position. D)The curve would be wider, but the vertex would be in the same position.

Answers

Answer:

Option (D)

Step-by-step explanation:

If a function f(x) is represented on a graph and we follow the transformation as,

f(x) → $k.f(x)$

1). If k ≥ 1, graph of the parent function f(x) will be stretched vertically by a factor k.

That means the transformed graph will be narrower.

2). If 0 < k < 1, graph will be vertically compressed or the transformed function will show a wider graph.

Following this rule,

f(x) → $(1)/(2).f(x)$ shows k = $(1)/(2)$ [Since 0 < $(1)/(2)$ < 1]

Therefore, transformed form will be wider.

Option (D) will be the answer.

Suppose you deposit $5,000 in a savings account that earns 3% annual interest. If you make no other withdrawals or deposits, how many years will it take the account balance to reach at least $6,000? A. 10 years. B. 6 years. C. 7 years. D. 4 years

Answers

You need to use the equation y=p(1+r)^t where p= the principle (initial amount), r= the rate of decay or growth, and t= the time. So y=5,000(1.03)^x is the equation you'd want to use. At 6 years you'd have $5,970.3 and at 7 $6,149.40, so the answer is C.

taking into consideration that the interest is compound (yearly)
the amount of money gather through the years can be calculated by

A = P (1+r)^(t)
6000 = 5000 (1.03)^t

t = ln(6000/5000)/ln(1.03) = 6.16 ≈ 7

c. 7 years

What ways can you collect 82 cents ? Using 13 coins .