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zalisa [80]
3 years ago
15

Find the surface area of a rectangular prism with base 4 m by 5 m and height 8 m.

Mathematics
2 answers:
Annette [7]3 years ago
6 0

Answer:

184

Step-by-step explanation:

A=2(wl+hl+hw)

plug in the numbers into the formula

Andre45 [30]3 years ago
4 0

Answer:

Step-by-step explanation:

surface area=area of sides +area of top and bottom

=2(4+5)×8+2×(4×5)

=2[(9×8)+20]

=2[72+20]

=2×92

=184 m²

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The answer to that question is 80.5
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3 years ago
2. Priscilla is drawing a map of her school that uses a scale in which 1.5 cm 1 point
lesantik [10]

Answer:

The answer is 60cm. (b)

Step-by-step explanation:

For every 1.5cm on the map, it represents 3 ft on real-size scale.

In other words, for every 1cm on the map, it represents 3/1.5 = 2ft on real size scale.

If you want to find 120ft of real-size scale, you can use 120ft/2ft * 1cm which gives 60cm

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3 years ago
Evaluate the Riemann sum for f(x) = 3 - 1/2 times x between 2 and 14 where the endpoints are included with six subintervals taki
Digiron [165]

Answer:

-6

Step-by-step explanation:

Given that :

we are to evaluate the Riemann sum for f(x) = 3 - \dfrac{1}{2}x from 2 ≤ x ≤ 14

where the endpoints are included with six subintervals, taking the sample points to be the left endpoints.

The Riemann sum can be computed as follows:

L_6 = \int ^{14}_{2}3- \dfrac{1}{2}x \dx = \lim_{n \to \infty} \sum \limits ^6 _{i=1} \ f (x_i -1) \Delta x

where:

\Delta x = \dfrac{b-a}{a}

a = 2

b =14

n = 6

∴

\Delta x = \dfrac{14-2}{6}

\Delta x = \dfrac{12}{6}

\Delta x =2

Hence;

x_0 = 2 \\ \\  x_1 = 2+2 =4\\ \\  x_2 = 2 + 2(2) \\ \\  x_i = 2 + 2i

Here, we are  using left end-points, then:

x_i-1 = 2+ 2(i-1)

Replacing it into Riemann equation;

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} \begin {pmatrix}3 - \dfrac{1}{2} \begin {pmatrix}  2+2 (i-1)  \end {pmatrix} \end {pmatrix}2

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 - (2+2(i-1))

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 - (2+2i-2)

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 -2i

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 -   \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 2i

L_6 =  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} 6 - 2  \lim_{n \to \infty}  \sum \imits ^{6}_{i=1} i

Estimating the integrals, we have :

= 6n - 2 ( \dfrac{n(n-1)}{2})

= 6n - n(n+1)

replacing thevalue of n = 6 (i.e the sub interval number), we have:

= 6(6) - 6(6+1)

= 36 - 36 -6

= -6

5 0
3 years ago
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Answer:

<u>1/4</u>

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3 0
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geniusboy [140]
1700÷40= 42.5 

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step two: you find the closest number that goes into 1700 which is 4 so 40×4= 160
step three: then you subtract 1700 from 160 and you'll get 100
step four: then you'll find the closest number that goes into a 100 and that will be 2 so 40×2=80
step 5: you'll then subtract 100 from 80 and you'll get 20 
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7 0
3 years ago
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