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allsm [11]
3 years ago
10

Alan has two more than twice as many chocolates as does alice, and half as many chocolates as does nadia. if alice has ‘a' numbe

r of chocolates, then in terms of ‘a', how many chocolates do alan, alice and nadia have
Mathematics
1 answer:
Misha Larkins [42]3 years ago
5 0
Alan has two more than twice as many chocolates as does alice
Alan = 2Alice + 2 ----(1)

Alan has half as many chocolates as does nadia
Alan = 1/2 Nadia
so Nadia = 2Alan ----(2)

if Alice has 'a' number of chocolates
from (1)
Alan = 2a + 2 ----(3)
from (2) and (3)
Nadia = 2 (2a+2) = 4a+4 ----(4)

So Alan Alice and Nadia have
(2a+2) + a + (4a+4) = 7a+ 6 #
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Two similar pyramids have base areas of 12.2 cm2 and 16 cm2. The surface area of the larger pyramid is 56 cm2. What is the surfa
docker41 [41]

Given =

Two similar pyramid have base area of 12.2 cm² and 16 cm².

surface area of the larger pyramid = 56 cm²

find out the surface area of the smaller pyramid

To proof =

Let us assume that the surface  area of the smaller pyramid be x.

as surface area of the larger pyramid is 56 cm²

Two similar pyramid have base area of 12.2 cm² and 16 cm².

by using ratio and proportion

we have

ratio of the base area of the pyramids : ratio of the surface area  of the pyramids

\frac{12.2}{16} : \frac{x}{56}

x = 12.2 ×56×\frac{1}{16}

by solvingthe above terms

we get

x =42.7cm²

Hence the surface area of the smaller pyramid be 42.7cm²

Hence proved










3 0
3 years ago
Read 2 more answers
Pls help<br><br> Given: △ABC, CM⊥ AB, BC = 5, AB = 7<br> CA = 4 sqrt(2)<br> Find: CM
babymother [125]

Answer:

The general plan is to find BM and from that CM. You need 2 equations to do that.

Step One

Set up the two equations.

(7 - BM)^2 + CM^2 = (4*sqrt(2) ) ^ 2 = 32

BM^2 + CM^2 = 5^2 = 25

Step Two

Subtract the two equations.

(7 - BM)^2 + CM^2  = 32

BM^2 + CM^2         = 25

(7 - BM)^2 - BM^2 = 7               (3)

Step three

Expand the left side of the new equation labeled (3)

49 - 14BM + BM^2 - BM^2 = 7    

Step 4

Simplify And Solve

49 - 14BM = 7              Subtract 49 from both sides.

-49 - 14BM = 7 - 49

- 14BM = - 42              Divide by - 14

BM = -42 / - 14

BM = 3

Step  Five

Find CM

CM^2 + BM^2 = 5^2

CM^2 + 3^2 = 5^2        Subtract 3^2 from both sides.

CM^2 = 25 - 9            

CM^2 = 16                     Take the square root of both sides.        

sqrt(CM^2) = sqrt(16)

CM = 4    < Answer

Step-by-step explanation:

6 0
2 years ago
Please help! i literally suck at geometry
sdas [7]
Don’t say like that!Think you can and you will achieve! :) :)
8 0
2 years ago
Given the diagram below. If G is the circumcenter of ABC. DC = 16, GB = 21 and AZ = 19, find each measure.​
Jlenok [28]

Answer:

Step-by-step explanation:

Properties of a circumcenter;

1). Circumcenter of a triangle is a point which is equidistant from all vertices.

2). Point where perpendicular bisectors of the sides of a triangle meet is called circumcenter of the triangle.

From the picture attached,

9). AG = GB = GC = 21

10). BC = 2(DC)

            = 2×16

            = 32

11). By applying Pythagoras Theorem in ΔGFB,

     GB² = GF² + FB²

     (21)² = GF² + (19)²

     441 = GF² + 361

     GF² = 441 - 361

     GF = \sqrt{80}

     GF = 8.9

12). By applying Pythagoras theorem in ΔGDB,

     GB² = DG² + BD²

     (21)² = (DG)² + (16)² [BD = DC = 16]

     DG² = 441 - 256

     DG = √185

     DG =  13.6

7 0
3 years ago
Write x^2 − 8x + 13 = 0 in the form (x −
padilas [110]
If you would like to write the equation in the above form, you can calculate this using the following steps:

x^2 - 8x + 13 = 0
(x - 4)^2 = x^2 - 8x + 16
(x - 4)^2 - 3 = 0
(x - 4<span>)^2 = 3
</span>
The correct result would be (x - 4<span>)^2 = 3.</span>
4 0
3 years ago
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