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ad-work [718]
3 years ago
12

Elsie, who loves to garden, has the garden plot above. The width of her current garden is 12 the length. Elsie wants to increase

both the width and length of her garden to twice the size. What is the area of the original garden? What is the area of the new garden? How do the areas compare?

Mathematics
1 answer:
pogonyaev3 years ago
4 0

Answer:

14*7=98    (Original area)

28*14=392      (New area)

The area of the new one is 4 times the original, since the width and length wee doubled.

28 is the doubled amount of 14. 14 is the doubled amount of 7.

Please but brainliest if you can.

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A man looked in his wallet and found he had 3 more $5 bills than $10 dollar bills. He had a total of $75. How many of each kind
Neko [114]

Answer:

Four 10s and seven 5s

Step-by-step explanation:

If you start with one 10, you have to have four 5s. After that you add one of each until you reach 75

3 0
3 years ago
A circular plate has circumference 16.3 inches. What is the area of this​ plate? Use 3.14 for pi.
serg [7]

Answer:

C = 2πr = 16.3

2(3.14)r = 16.3

r = 16.3/6.28

r = 2.5955414

Area:  A = πr2

A = 3.14(2.5955414)2

A = 21.15366242 in^{2}

Step-by-step explanation:

The answer would be 21.2 in^{2}

6 0
3 years ago
Rationalize the denominator and simplify.
cestrela7 [59]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3151018

——————————
 
     \mathsf{\dfrac{3\sqrt{6}+5\sqrt{2}}{4\sqrt{6}-3\sqrt{2}}}


Multiply and divide by the conjugate of the denominator, which is  \mathsf{4\sqrt{6}+3\sqrt{2}:}

     \mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}}


Expand those products and eliminate the brackets:
 
     \mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(3\sqrt{6}+5\sqrt{2}\big)\cdot 3\sqrt{2}}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(4\sqrt{6}-3\sqrt{2}\big)\cdot 3\sqrt{2}}}\\\\\\ \mathsf{=\dfrac{12\cdot \big(\sqrt{6}\big)^2+20\sqrt{2}\cdot \sqrt{6}+9\cdot \sqrt{6}\cdot \sqrt{2}+15\cdot \big(\sqrt{2}\big)^2}{16\cdot \big(\sqrt{6}\big)^2-12\cdot \sqrt{2}\cdot \sqrt{6}+ 12\cdot \sqrt{6}\cdot \sqrt{2}-9\cdot \big(\sqrt{2}\big)^2}}\\\\\\ \mathsf{=\dfrac{12\cdot 6+20\sqrt{2\cdot 6}+9\sqrt{6\cdot 2}+15\cdot 2}{16\cdot 6-12\sqrt{2\cdot 6}+ 12\sqrt{6\cdot 2}-9\cdot 2}}\\\\\\ \mathsf{=\dfrac{72+20\sqrt{12}+9\sqrt{12}+30}{96-12\sqrt{12}+12\sqrt{12}-18}}

     \mathsf{=\dfrac{72+29\sqrt{12}+30}{96-18}}\\\\\\ \mathsf{=\dfrac{102+29\sqrt{2^2\cdot 3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot \sqrt{2^2}\cdot \sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot 2\sqrt{3}}{78}} 

     \mathsf{=\dfrac{102+58\sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{\,\diagup\!\!\!\! 2\cdot \big(51+29\sqrt{3}\big)}{\diagup\!\!\!\! 2\cdot 39}} 

     \mathsf{=\dfrac{51+29\sqrt{3}}{39}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)

4 0
3 years ago
In the diagram below, what needs to be labeled in order to prove triangle AHL and triangle EML are congruent by the SSS postulat
Kobotan [32]

Answer:

HL = LM

Step-by-step explanation:

We have been given a diagram in which sides AH= EM and AL=EL.

Now we need to find which statement is missing to prove that triangles AHL and EML are congruent using SSS.

SSS means (side - side - side)

Two pair of equal sides are already given so we have completed SS of SSS.

Last "S" means we need one more pair of equal sides so that must be using remaining sides HL and LM

Hence final answer is HL = LM.

4 0
3 years ago
The graph below shows the rate that water leaves a tank, in gallons per hour, where t is measured in hours. (Please give exact a
defon

Answer:

Part a) 10 gallons

Part b) 4.59 hours

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

Part a) How much water left the tank during the 6 hours shown?

To find out how much water left the tank during the 6 hours, determine the area of the graph in that interval

so

Calculate the area of the rectangle plus the area of the triangle

(4)(2)+\frac{1}{2}(6-4)(2)

8+2=10\ gal

Part b) How many hours did it take for 9 gallons to leave the tank?

we know that

In the interval [0,4]

8 gallons of water leaves the tank in 4 hours (remember that the area in that interval is equal to 8 gallons)

so

Find out in the interval (4,6] how  many hours did it take for 1 gallon to leave the tank

First determine the equation of the line in the interval (4,6)

we have the points (4,2) and (6,0)

The slope is equal to

m=(6-2)/(0-4)=-1

The equation of the line is

y=-(x-6)\\y=-x+6

Determine the area of graph in the interval (4,x) (Is the area of trapezoid)

so

A=\frac{1}{2}(2+(-x+6))((x-4)

A=\frac{1}{2}(8-x)((x-4)

A=\frac{1}{2}(-x^2+12x-32)\\\\A=-\frac{1}{2}x^{2}+6x-16

For A=1 gal

1=-\frac{1}{2}x^{2}+6x-16

-\frac{1}{2}x^{2}+6x-17=0

solve the quadratic equation by graphing

The solution is x=4.59 hours

see the attached figure N 2

3 0
3 years ago
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