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3 years ago

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14. A=lwh , solve for w

1 answer:

notka56 [123]3 years ago

6 0

A/lh=w is the correct answer

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Fiona Follies is a small bakery that sells cupcakes and muffins. A cupcake sells for $3 and a muffin sells for $2. When all rela

Elan Coil [88]

Note that fiona cannot sell decimal number of cupcakes or muffins. That's why the options A and D are false.

Consider all remaining options:

B. 40 cupcakes and 80 muffins will cost $(40·3+80·2)=$(120+160)=$280, but Fiona Follies sold at least $300 worth of cupcakes and muffins. Thus, this option is false.

C. 60 cupcakes and 70 muffins will cost $(60·3+70·2)=$(180+140)=$320. Now consider expenses. $(60·0.75+70·0.5)=$(45+35)=$80. This option is true.

E. 80 cupcakes and 80 muffins will cost $(80·3+80·2)=$(240+160)=$400. Now consider expenses. $(80·0.75+80·0.5)=$(60+40)=$100 (exactly $100). This option is true.

Answer: correct options are C and E.

4 0

3 years ago

Find the surface area of the cylinder to the nearest tenth of a square unit. Use 3.14 for pi :))) thank you !!

Leni [432]

Answer:

390.8 ft2 I think

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

I will give u 100 points if u can answer this quition pls it is hard for me John has three pieces of rope. One piece was 31 cm l

serious [3.7K]

Answer:

Add 0.31 + 0.33 + 0.45;

Step-by-step explanation:

Hope I got it right! If not i'm sorry.

7 0

3 years ago

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Given your sister's incomplete text,

egoroff_w [7]

Explanation:

Quadrants 2 and 3 houses points that have negative x-values, while as Quadrants 1 and 4 houses points that have positive x-values.

If 'your sister' completes the text with negative y-value, the point would be in Q3.

If 'your sister' completes the text with positive y-value, the point would be in Q2.

8 0

3 years ago

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Estimate 1 13 cos(x2) dx 0 using the Trapezoidal Rule and the Midpoint Rule, each with n = 4. (Round your answers to six decimal

babunello [35]

Answer:

(a) 4.152698

(b) 3.215557

Step-by-step explanation:

(a)

$\int\limits^{13}_1 {cos(x^2)} \, dx =M_n=$\sum_{n=1}^{\infty} f(m_i)\Delta x $$

n=4, so :

Each subinterval has length :

$\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3$

Therefore the subintervals consist of:

$[1,5], [5,9], [9,13]$

Now, the midpoints of these subintervals are:

$\frac{1+5}{2} =3\\\\\frac{5+9}{2} =7\\\\\frac{9+13}{2} =11$

Hence:

$M_4= 3*(cos(3^2))+3*(cos(7^2))+3*cos((11^2))\approx 4.152698$

(b)

$\int\limits^{13}_1 {cos(x^2)} \, dx =T_n=\frac{\Delta x}{2} (f(x_o)+2f(x_1)+2f(x_2)+...+2f(x_n_-_1)+f(x_n))$

n=4, so :

Each subinterval has length :

$\Delta x= \frac{b-a}{n} =\frac{13-1}{4} =\frac{12}{4} =3$

Therefore the subintervals consist of:

[1,5], [5,9], [9,13]

The endpoints of the subintervals consist of:

5,9

Hence:

$T_4= \frac{3}{2}(cos(1^2)+2*cos(5^2)+2*cos(9^2)+cos(13^2)) \approx 3.215557$

8 0

2 years ago