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tensa zangetsu [6.8K]
3 years ago
14

It would be muchly appreciated

Mathematics
1 answer:
vovangra [49]3 years ago
4 0

Answer:

g= -32

Step-by-step explanation:

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What is the ratio of yellow butterflies to total butterflies? Choose the correct option
larisa86 [58]

Answer:2/3

Step-by-step explanation: there is 2 yellow out of 3 butterfiles

6 0
3 years ago
Kiara made a square baby blanket she decided to add one foot of material to one side then cut 4 inches of material off from the
schepotkina [342]

Answer:

Step-by-step explanation:

960 which is the area u divide by the one foot which is 144 square units to get your answer,6.66666666667

8 0
3 years ago
A farmer wants to fence in a rectangular plot of land adjacent to the north wall of his barn. No fencing is needed along the bar
Dmitriy789 [7]

Answer:

Width  = 166.7 ft

Length = 250 ft.

Step-by-step explanation:

From the diagram at the ATTACHMENT

Let us denote the the length of the rectangular area as "x" and

the width of the rectangular area as "y"

Since, cost of the fence on the western side is shared with a neighbor, we can say  it is $7 per foot. Otherwise, it is $14 per foot. The northern length is not fenced.

From the question, we know that the  cost is not to exceed $7000, therefore we can expressed it as

7y + 14x + 14y = 7000

14x + 21y = 7000

Divide through by 14, we have

(x + 1.5) = 500  ------------- eqn (*)            

But the area of the fenced

= A = (x×y)      -------------- eqn  (**)                    

Substitute (*) into (**) we have

A = (500 - 1.5y)*y

= 500y - 1.5y²

In other to maximize the Area, we can find the differential of A

A'(y) = 500 - 3y

At A'(y) = 0.

500-3y= 0

3y=500

y= 166.67 ft

From eqn(*)

(x+ 1.5) = 500

x = 500 - 1.5y

500-1.5(166.67)

= 250ft

Therefore, the dimensions for the plot that would enclose the most area are

Width  of 166.67 ft and Length = 250 ft.

4 0
4 years ago
What is 0.433 rounded to the nearest tenths
Fynjy0 [20]
0.433 is already rounded to the hundredths so to round it to the nearest tenths, remove the last number.

Your answer is 0.43
6 0
3 years ago
Read 2 more answers
Please help me solve this problem ASAP
DiKsa [7]

\bold{\huge{\blue{\underline{ Solution }}}}

<h3><u>Given </u><u>:</u><u>-</u></h3>

  • <u>The </u><u>right </u><u>angled </u><u>below </u><u>is </u><u>formed </u><u>by </u><u>3</u><u> </u><u>squares </u><u>A</u><u>, </u><u> </u><u>B </u><u>and </u><u>C</u>
  • <u>The </u><u>area </u><u>of </u><u>square </u><u>B</u><u> </u><u>has </u><u>an </u><u>area </u><u>of </u><u>1</u><u>4</u><u>4</u><u> </u><u>inches </u><u>²</u>
  • <u>The </u><u>area </u><u>of </u><u>square </u><u>C </u><u>has </u><u>an </u><u>of </u><u>1</u><u>6</u><u>9</u><u> </u><u>inches </u><u>²</u>

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>area </u><u>of </u><u>square </u><u>A</u><u>? </u>

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u><u> </u></h3>

The right angled triangle is formed by 3 squares

<u>We </u><u>have</u><u>, </u>

  • Area of square B is 144 inches²
  • Area of square C is 169 inches²

<u>We </u><u>know </u><u>that</u><u>, </u>

\bold{ Area \: of \: square =  Side × Side }

Let the side of square B be x

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ 144 =  x × x }

\sf{ 144 =  x² }

\sf{ x = √144}

\bold{\red{ x = 12\: inches }}

Thus, The dimension of square B is 12 inches

<h3><u>Now, </u></h3>

Area of square C = 169 inches

Let the side of square C be y

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ 169 =  y × y }

\sf{ 169 =  y² }

\sf{ y = √169}

\bold{\green{ y = 13\: inches }}

Thus, The dimension of square C is 13 inches.

<h3><u>Now, </u></h3>

It is mentioned in the question that, the right angled triangle is formed by 3 squares

The dimensions of square be is x and y

Let the dimensions of square A be z

<h3><u>Therefore</u><u>, </u><u>By </u><u>using </u><u>Pythagoras </u><u>theorem</u><u>, </u></h3>

  • <u>The </u><u>sum </u><u>of </u><u>squares </u><u>of </u><u>base </u><u>and </u><u>perpendicular </u><u>height </u><u>equal </u><u>to </u><u>the </u><u>square </u><u>of </u><u>hypotenuse </u>

<u>That </u><u>is</u><u>, </u>

\bold{\pink{ (Perpendicular)² + (Base)² = (Hypotenuse)² }}

<u>Here</u><u>, </u>

  • Base = x = 12 inches
  • Perpendicular = z
  • Hypotenuse = y = 13 inches

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ (z)² + (x)² = (y)² }

\sf{ (z)² + (12)² = (169)² }

\sf{ (z)² + 144 = 169}

\sf{ (z)² = 169 - 144 }

\sf{ (z)² = 25}

\bold{\blue{ z = 5 }}

Thus, The dimensions of square A is 5 inches

<h3><u>Therefore</u><u>,</u></h3>

Area of square

\sf{ = Side × Side }

\sf{ = 5 × 5  }

\bold{\orange{ = 25\: inches }}

Hence, The area of square A is 25 inches.

6 0
3 years ago
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