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GaryK [48]
3 years ago
12

#5 please! Need help

Mathematics
1 answer:
butalik [34]3 years ago
3 0

Answer:

Step-by-step explanation:

Distance = Speed x Time and you need speed so the equation is

D divided by T = S

Cutwright   35 laps divided by 84 min .... 42mph

Evans   42 laps divided by 96.6 min ... 43 mph

Loza 38 laps divided by 102.6 min ... 37 mph

Evans went slightly faster than Cutwright

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What is the area of this figure?<br><br><br> 28 yd²<br><br> 40 yd²<br><br> 52 yd²<br><br> 64 yd²
Lubov Fominskaja [6]

Answer:  A= s^2 <- for a square

A= 1/2*b*h <- for a triangle

A= 4^2= 16

A= 1/2*8*6 (The whole height is 8 yd)

A= 1/2*48

A= 24

24+16= 40 yd^2

"B" is the answer.

I hope this helps!

Step-by-step explanation: A= s^2 <- for a square

A= 1/2*b*h <- for a triangle

A= 4^2= 16

A= 1/2*8*6 (The whole height is 8 yd)

A= 1/2*48

A= 24

24+16= 40 yd^2

"B" is the answer.

I hope this helps!

6 0
3 years ago
Mr.kiniston jogged how many miles
Dimas [21]
The answer is letter G.
4 0
3 years ago
Read 2 more answers
Find the area of the following<br> kite:<br> A = [?] m²<br> 40 m<br> 16 m<br> 16 m<br> 6 m
Rama09 [41]

Answer:

Area_{kite}=736m^2

Step-by-step explanation:

There are a few methods to find the area of this figure:

1. kite area formula

2. 2 triangles (one top, one bottom)

3. 2 triangles (one left, one right)

4. 4 separate right triangles.

<h3><u>Option 1:  The kite area formula</u></h3>

Recall the formula for area of a kite:  Area_{kite}=\frac{1}{2} d_{1}d_{2} where d1 and d2 are the lengths of the diagonals of the kite ("diagonals" are segments that connect non-adjacent vertices -- in a quadrilateral, vertices that are across from each other).

If you've forgotten why that is the formula for the area of a kite, observe the attached diagram: note that the kite (shaded in) is half of the area of the rectangle that surrounds the kite (visualize the 4 smaller rectangles, and observe that the shaded portion is half of each, and thus the area of the kite is half the area of the large rectangle).

The area of a rectangle is Area_{rectangle}=bh, sometimes written as Area_{rectangle}=bh, where w is the width, and h is the height of the rectangle.

In the diagram, notice that the width and height are each just the diagonals of the kite.  So, the <u>Area of the kite</u> is <u>half of the area of that surrounding rectangle</u> ... the rectangle with sides the lengths of the kite's diagonals.Hence, Area_{kite}=\frac{1}{2} d_{1}d_{2}

For our situation, each of the diagonals is already broken up into two parts from the intersection of the diagonals.  To find the full length of the diagonal, add each part together:

For the horizontal diagonal (which I'll call d1): d_{1}=40m+6m=46m

For the vertical diagonal (which I'll call d2): d_{2}=16m+16m=32m

Substituting back into the formula for the area of a kite:

Area_{kite}=\frac{1}{2} d_{1}d_{2}\\Area_{kite}=\frac{1}{2} (46m)(32m)\\Area_{kite}=736m^2

<h3><u /></h3><h3><u>Option 2:  The sum of the parts (version 1)</u></h3>

If one doesn't remember the formula for the area of a kite, and can't remember how to build it, the given shape could be visualized as 2 separate triangles, the given shape could be visualized as 2 separate triangles (one on top; one on bottom).

Visualizing it in this way produces two congruent triangles.  Since the upper and lower triangles are congruent, they have the same area, and thus the area of the kite is double the area of the upper triangle.

Recall the formula for area of a triangle:  Area_{triangle}=\frac{1}{2} bh where b is the base of a triangle, and h is the height of the triangle <em>(length of a perpendicular line segment between a point on the line containing the base, and the non-colinear vertex)</em>.  Since all kites have diagonals that are perpendicular to each other (as already indicated in the diagram), the height is already given (16m).

The base of the upper triangle, is the sum of the two segments that compose it:  b=40m+6m=46m

<u>Finding the Area of the upper triangle</u>Area_{\text{upper }triangle}=\frac{1}{2} (46m)(16m) = 368m^2

<u>Finding the Area of the kite</u>

Area_{kite}=2*(368m^2)

Area_{kite}=736m^2

<h3><u>Option 3:  The sum of the parts (version 2)</u></h3>

The given shape could be visualized as 2 separate triangles (one on the left; one on the right).  Each triangle has its own area, and the sum of both triangle areas is the area of the kite.

<em>Note:  In this visualization, the two triangles are not congruent, so it is not possible to  double one of their areas to find the area of the kite.</em>

The base of the left triangle is the vertical line segment the is the vertical diagonal of the kite.  We'll need to add together the two segments that compose it:  b=16m+16m=32m.  This is also the base of the triangle on the right.

<u>Finding the Area of left and right triangles</u>

Area_{\text{left }triangle}=\frac{1}{2} (32m)(40m) = 640m^2

The base of the right triangle is the same length as the left triangle: Area_{\text{right }triangle}=\frac{1}{2} (32m)(6m) = 96m^2

<u>Finding the Area of the kite</u>

Area_{kite}=(640m^2)+(96m^2)

Area_{kite}=736m^2

<h3><u>Option 4:  The sum of the parts (version 3)</u></h3>

If you don't happen to see those composite triangles from option 2 or 3 when you're working this out on a particular problem, the given shape could be visualized as 4 separate right triangles, and we're still given enough information in this problem to solve it this way.

<u>Calculating the area of the 4 right triangles</u>

Area_{\text{upper left }triangle}=\frac{1}{2} (40m)(16m) = 320m^2

Area_{\text{upper right }triangle}=\frac{1}{2} (6m)(16m) = 48m^2

Area_{\text{lower left }triangle}=\frac{1}{2} (40m)(16m) = 320m^2

Area_{\text{lower right }triangle}=\frac{1}{2} (6m)(16m) = 48m^2

<u>Calculating the area of the kite</u>

Area_{kite}=(320m^2)+(48m^2)+(320m^2)+(48m^2)

Area_{kite}=736m^2

8 0
2 years ago
The newborn weights 7 pounds and is 21 inches long. What is the ratio of weight to length?​
Nikolay [14]

Answer:

1:3

Step-by-step explanation:

The required ratio will be ,

7:21

= 1 :3

4 0
3 years ago
Read 2 more answers
WILL MARK AS BRAINLIEST FOR THESE QUESTIONS.
laiz [17]

1.

Let the sandwich be represented by = x

Let the drink be represented by = y

As given, Joe charges $10.00 for two sandwiches and one drink, so equation becomes 2x+y=10    ....(i)

He charges $6.50 for one sandwich and one drink, so equation becomes x+y=6.50  ....(ii)

Subtracting (ii) from (i) we get

x=3.50

And as x+y=6.50

y=6.50-3.50=3.00

Hence, cost of a sandwich is $3.50 and a drink is $3.00

2.

the question is incomplete. coordinates are missing

3.

Let x represent the number of performances

cost = 8000+300x

revenue = 700x

To break even, the cost must be equal to the revenue.

8000+300x=700x

400x=8000

so x=20

Hence, 20 performances are needed to break even.

4.

Let us take x for 20% material, y for 50% material

\frac{20x+50y}{200}=30

20x+50y=6000    ......... (1)

Accounting for volumes to use, x+y=200 or x=200-y ................ (2)

Putting the value of x in (1)

20(200-y)+50y=6000

4000-20y+50y=6000

4000+30y=6000

30y=2000

y=66.67

x=200-y = 200-66.67 =133.33

Hence, 133.33 ml of 20% acetic acid and 66.67% of 50% acetic acid is needed.





3 0
3 years ago
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