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Lelechka [254]
2 years ago
8

PLS HELP Q1: Describe the pattern in the table:

Mathematics
1 answer:
Whitepunk [10]2 years ago
5 0

Answer:

The length is always 1.6 times larger than the width.

When going from each model: A to B to C to D, the Width has an added 5cm each time.

When going from each model: A to B to C to D, the Length has an added 8cm each time.

Step-by-step explanation:

8/5 = 1.6

16/10 = 1.6

24/15 = 1.6

32/20 = 1.6

The length is 1.6 times larger than the width.

When going from each model: A to B to C to D, the Width has an added 5cm each time.

When going from each model: A to B to C to D, the Length has an added 8cm each time.

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The first five terms of a pattern are shown.
Andrej [43]

Answer:

C

Step-by-step explanation:

There is a common ratio between consecutive term in the sequence, that is

\frac{3}{10} ÷ \frac{3}{100} = 3 ÷ \frac{3}{10} = 30 ÷ 3 = 300 ÷ 30 = 10

This indicates the sequence is geometric with n th term

a_{n} = a(r)^{n-1}

where a is the first term and r the common ratio

Here a = \frac{3}{100} and r = 10 , thus

a_{n} = \frac{3}{100} (10)^{n-1}

    = \frac{3}{10^{2} } × 10^{n-1}

    = 3 × 10^{-2} × 10^{n-1}

    = 3 × 10^{n-3}

Thus

a_{n} = 3(10)^{n-3} → C

6 0
3 years ago
WILL GIVE BRAILIEST TO THE BEST ANSWER
xxTIMURxx [149]
Answer is C.

12 servings        9 servings
----------------- =  ---------------
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4 0
3 years ago
Can anyone please help!!!
quester [9]

Answer:

Question 9

(a) The distance, Jalaj walks in one day is 4.4 km

(b) The amount Jalaj raises after walking for 22 km at the end of the 5 days is $8

Question 10

(b) The difference between the largest and smallest areas is 2,400 m²

Step-by-step explanation:

Question 9

(a) The distance Jalaj walks in 5 days = 22 km

Whereby Jalaj walks equal distance every day, we have;

The distance, Jalaj walks in one day = 22 km/5 days = 4.4 km/day

The distance, Jalaj walks in one day = 4.4 km

(b) The amount he raises for every kilometer he walks = $1.60

The amount he raises after walking for 22 km at the end of the 5 days = 5 × $1.60  = $8

Question 10

(b) The given side length of the square = 120 meters to the nearest 10 meters

Therefore;

The maximum dimension for the side length of the square = 120 + 10/2 = 125

The largest possible area of the square, A_l = 125 m × 125 m = 15,625 m²

The minimum dimension for the side length of the square = 120 m - 10 m/2 = 115 m

The smallest possible area of the square, A_s = 115 m × 115 m = 13,225 m².

The difference between the largest and smallest areas, A_l - A_s = 15,625 - 13,225 = 2,400 m².

8 0
2 years ago
Options 1,2,3 or4?<br><br> Help!<br> Last one I swear!
lapo4ka [179]

Answer:

option 3 = 3.75

Step-by-step explanation:

x is 3.75

.

8 0
3 years ago
Read 2 more answers
A box is being created out of a 15 inch by 10 inch sheet of metal. Equal-sized squares are cutout of the corners, then the sides
ivolga24 [154]

Answer:

Therefore the dimensions of the square should be 0.1528 inch by 0.1528 inch so, the box  has largest volume.

Step-by-step explanation:

Given that,

A box is being created out of a 15 inches by 10 inches sheet of metal.

The length of the one side of the squares which are cut out of the each corners of the metal sheet be x.

The length of the metal box be = (15-2x) inches.

The width of the metal box be =(10-2x) inches

The height of the metal box be =x inches

Then, the volume of the metal box= length×width×height

                                                         =(15-2x)(10-2x)x cubic inches

                                                         =(150x-50x²+4x³) cubic inches

∴ V= 4x³-50x²+15x

Differentiating with respect to x

V'=12x²-100x+15

Again differentiating with respect to x

V''=24x-100

For maximum or minimum value, V'=0

12x²-100x+15=0

Apply quadratic formula x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}, here a=12, b= -100 and c=15

x=\frac{-(-100)\pm\sqrt{(-100)^2-4.12.15}}{2.12}

\Rightarrow x=\frac{100\pm\sqrt{9280}}{2.12}

\Rightarrow x=0.1528,8.18

For x= 8.18, The value of (15-2x) and (10-2x) will negative.

∴x=0.1528 .

Now, V''|_{x=0.1528}=24(0.1528)-100

∴At x=0.1528 inch, the volume of the metal box will be maximum.

Therefore the dimensions of the square should be 0.1528 inch by 0.1528 inch so, the box  has largest volume.

4 0
3 years ago
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