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kiruha [24]
3 years ago
10

HELPPPP FOR 30 points!!! HELPP

Mathematics
1 answer:
lara [203]3 years ago
3 0

Answer: 1. $499

2. $270.48

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i will say you have the right answer selected in the photo


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100 miles

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Help me solve (b) in this quadrilateral
Savatey [412]

Answer:

b = 87°

Step-by-step explanation:

In order to answer this question, we need to utilise an important angle fact which is <em>angles in a quadrilateral add up to 360° </em>

Using the information we can set up an equation to find the value of b

→ Form equation

63 + 140 + 70 + b = 360

→ Simplify

273 + b = 360

→ Minus 273 from both sides isolate b

b = 87°

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3 years ago
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Step-by-step explanation:

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2 years ago
The top and bottom margins of a poster are each $3$ cm and the side margins are each $2$ cm. If the area of printed material on
babymother [125]

Answer:

Therefore the dimension of the poster is 12 cm by 8 cm.

Step-by-step explanation:

Let the length of the poster be x and the width be y.

Given that the area of the poster is 96 cm².

∴xy =96

\Rightarrow y= \frac{96}{x}

The sides margins each are 2 cm and the top and bottom margins of the poster are each 3 cm.

The length of printing space is =(x- 2.3) cm

                                                   = (x-6) cm

The width of the printing space is =(y-2.2) cm

                                                         =( y-4 )cm

The area of the printing space is A=(x-6)(y-4) cm²

∴A=(x-6)(y-4)  

\Rightarrow A=(x-6)(\frac{96}{x}-4)    [ Putting y= \frac{96}{x}  ]

\Rightarrow A=120-\frac{576}{x}-4x

Differentiating with respect to x

A'= \frac{576}{x^2}-4

Again differentiating with respect to x

A''=-\frac{1152}{x^3}

To find the minimum area, we set A'=0

\therefore  \frac{576}{x^2}-4=0

\Rightarrow \frac{576}{x^2}=4

\Rightarrow x^2=\frac{576}{4}

\Rightarrow x^2 =144

\Rightarrow x=\pm 12

Dimension can't be negative.

Therefore x=12

If x=12, the value of A''>0,then at x=12, the area of the poster will be minimum.

If x=12, the value of A''<0,then at x=12, the area of the poster will be minimum.

\therefore A''|_{x=12}=-\frac{1152}{12^3}

Therefore at x= 12 cm the area of the poster will be maximum.

The width of the poster is y=\frac{96}{12} = 8 cm

Therefore the dimension of the poster is 12 cm by 8 cm.

3 0
3 years ago
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