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3 years ago

Why is graphing not always the best option for solving systems?

2 answers:

6 0

Best Answer: When graphing and locating the point of intersection & hence the solution, you might find that this point does not have integer co ordinates. You would have to estimate the approximate coordinates. Hence, the algebraic methods are preferred, as you could then state the exact co ordinates like (5/3, 4/9).

RideAnS [48]3 years ago

3 0

1, Might not be precise
2, Can be hard to plot the graphs
3, If there are many solutions that will be really hard to plot sometimes.

This is just from the top of my head.

You might be interested in

In the figure, Δ ABC Δ XYZ. What is the perimeter of ΔABC? Show your work.

slava [35]

Answer:

Step-by-step explanation:

∆ABC~ ∆XYZ

AB= 11 that is half of XY = 22

therefore XZ/2 = AC (base) = 32/2 = 16

and YZ /2 = BC(hypotenuse) = 44/2 = 22

(a= 11, b= 16, c= 22)

perimeter of a triangle

P = a+b+c

P= 11+16+22= 49

8 0

3 years ago

What is the volume of the cylinder? Round to the nearest 10th.

nata0808 [166]

Answer:169.7

Step-by-step explanation:

To find the volume of a cylinder you can use the math formula πr^2H where r is the radius and H is the height. If you plug in the values you get π3^2(6) simplify 3^2 to get (9π)(6), you can multiply 9 by 6 to get 54 then multiply that by 3.1415 to get 169.65 which rounds to 169.7

6 0

3 years ago

Find the volume of the cylinder height 10 in base radius 3 in

tino4ka555 [31]

Answer: 282.60in³

Step-by-step explanation:

When calculating the volume of a cylinder, the formula to use is:

= πr²h

where,

π = 3.14

r = radius = 3 in

h = height = 10 in

Volume = πr²h

= 3.14 × 3² × 10

= 282.60

Therefore, the volume of the cylinder is 282.60in³

3 0

2 years ago

Help pleaseeeeeeeeeeeeeeeeeeeeeee

bixtya [17]

Answer: $\bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}$

<u>Step-by-step explanation:</u>

(1) (12, 18, 27, ...)

The common ratio is:

$r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\ r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}$

$(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}$

The equation is:

$a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\ r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}$

3 0

2 years ago

There are 32 times students can work in the computer lab during one week. If 24 students can work in the computer lab at one tim

uysha [10]

600 students can work in the lab

4 0

2 years ago

Read 2 more answers