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Llana [10]
4 years ago
6

Algebraic 60- Y = 48 Y =

Mathematics
1 answer:
erica [24]4 years ago
3 0
Add y to both sides
60=48y+y
undistribute y
60=y(48+1)
60=y(49)
divide both sides by 49
60/49=y
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Solve c =(a² + 3b)/4 for b.<br><br>Bruv take all my points I literally blow at math :'(​
Sati [7]

Answer:

b=\frac{4c-a^{2} }{3}

Step-by-step explanation:

First: multiply both sides by 4. 4 times c is 4c and the other sides cancels out as you are doing the inverse operation of division

Then, you have 4c=a^{2} +3b

Subtract both sides by a squared

DO NOT TAKE THE SQUARE ROOT! This is because it is a squared plus 3b so you have to do the inverse of addition

From that you get 4c-a^{2} =3b

Finally, divide both sides by 3.

You get  b=\frac{4c-a^{2} }{3}

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3 years ago
Teresa Gonzalez and Alan Carillo spent a total of $215.75 on their prom
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Answer:$139.88

Step-by-step explanation:

215,75-75.87 is 139.88

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3 years ago
a school purchsed basketball for $20 each and footballs for $15 each. they a brought a total of 28 for $480 how many of each typ
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13

Step-by-step explanation:

3 0
4 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
Read 2 more answers
Find the shaded sector and the length of the arc. Round to the nearest hundredth. ​
Dovator [93]

Answer:

Step-by-step explanation:

I'll show you how to do the first one; the other are exactly the same, so pay attention.

The formula for arc length is

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AL=\frac{40}{360}*2(3.1415)(3). I'm going to reduce that fraction a bit (and I'll use the same reduction in the Area of a sector coming up next):

AL=\frac{1}{9}*2(3.1415)(3) which makes

AL = 2.09 units. Now for Area of the Sector. The formula is almost identical, but instead uses the idea that the area of a circle is πr²:

A_s=\frac{\theta}{360}*\pi r^2 where θ is, again, the measure of the central angle (which is the same as the measure of the arc it intercepts). Filling in:

A_s=\frac{1}{9}*(3.1415)(3)^2 which simplifies a bit to

A_s=\frac{1}{9}*(3.1415)(9). As you can see, the 9's cancel each other out, leaving you with

A_s=3.14 units²

7 0
3 years ago
Read 2 more answers
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