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3 years ago

If you are struggling with this question this is the answer

Mathematics

1 answer:

Blizzard [7]3 years ago

6 0

Answer: she is right the answer is 17.6

Step-by-step explanation:

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Will mark brainlyist if right please helpp!!

Len [333]

Answer:

Step-by-step explanation:

7 0

3 years ago

2.26-30 3x+y=6 (0,_) (_,0) (3,_) (6,_) (5,_)

bekas [8.4K]

Answer:

Step-by-step explanation:

In ordered pairs (a,b) a is the x value and b is a y value.

if we have 3x+y=6

if x=0, y=6 --> (0,6)

if y=0, x=2 -->(2,0)

if x=3, y=-3--> (3, -3)

if x=6, y= -12 --->(6, -12)

if x=6, y= -9 ---> (5, -9)

7 0

2 years ago

You have 20 coins in your pocket that are either quarter or dimes. They total $3.95. How many quarters do you have? A.16 B.13 C.

bekas [8.4K]

Answer:

B. 13

Step-by-step explanation:

Solve this by setting up system of equations:

0.25x + 0.10y = 3.95

x + y = 20

x equals the number of quarters and y equals the number of dimes. 0.25 is the value of a quarter and 0.10 is the value of a dime.

1. Multiply one equation to have the same coefficient as the other

0.10 · (x + y = 20) = 0.10x + 0.10y = 2

2. Subtract to find the value of one variable

0.25x + 0.10y = 3.95

- 0.10x + 0.10y = 2

0.15x = 1.95

3. Solve for x by dividing both sides by 0.15

x = 13

3 0

3 years ago

What is the slope of a line perpendicular to the line with equation y = 1/6x – 2? help pls very urgent

valina [46]

Solution:

1) Simplify \frac{1}{6}x to \frac{x}{6}
y=\frac{x}{6}-2

2) Add 2 to both sides
y+2=\frac{x}{6}

3) Multiply both sides by 6
(y+2)\times 6=x

4) Regroup terms
6(y+2)=x

5) Switch sides
x=6(y+2)

Done!

7 0

3 years ago

How to work it out logically? Question a !!!

Arturiano [62]

Answer:

Fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

Step-by-step explanation:

Given the figure with dimensions. we have to find the area of given figure.

Area of figure=ar(1)+ar(2)+ar(3)

Area of region 1 = ar(ANGI)+ar(AIB)

$=L\times B+\frac{1}{2}\times base\times height\\\\=[1500\times (5000-2000-1500)]+\frac{1}{2}\times (3000-1500)\times (5000-2000-1500)\\\\=3375000m^2=337.5ha$

Area of region 2 = ar(DHBC)

$=2000\times1500\\\\=3000000m^2=300ha$

Area of region 3 = ar(GFEH)

$(2000+1500)\times 1000\\\\=3500000m^2=350ha$

Hence, Area of figure=ar(1)+ar(2)+ar(3)=337.5ha+300ha+350ha

=987.5 ha

Now, we have to do straight-line fencing such that area become half and cost of fencing is minimum.

Let the fencing be done through x m downward from B which divides the two into equal area.

⇒ Area of upper part above fencing=Area of lower part below fencing

⇒ $ar(ANGB)+ar(GKLB)=ar(KLCM)+ar(MDCF)\\\\337500+3000x=(3500-x)\times 1000+2000(1500-x)\\\\3375000+3000x=3500000-1000x+3000000-2000x\\\\6000x=315000\\\\x=\frac{315000}{6000}=520.8m$

Hence, fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.

7 0

3 years ago

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