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mr Goodwill [35]
3 years ago
9

How can you tell if a equation is extraneous

Mathematics
1 answer:
geniusboy [140]3 years ago
6 0
Do you mean an extraneous solution? I will tell you what I think. To tell if a solution is extraneous you need to go back to the original problem and check if there actually is a solution or if the solution is actually right. If not then it is extraneous.
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Explain how the following diagram demonstrates the pythagorean theorem?
Yuliya22 [10]

Answer:

the largest block represents the hypotenuse which would be c² the smallest can either be a or b and the block thats left over will be the letter you didn't use for the small block so in terms the blocks represents a²+b²=c²

7 0
3 years ago
Read 2 more answers
arrange the circles (represented by their equations in general form) in ascending order of their radius lengths,  x^+y^-2x+2y-1=
Lubov Fominskaja [6]
From the order you gave me in the question the answer is equation number 
4 1 3 2 6 5
Those are the equation numbers in ascending order based on their radius lengths
7 0
3 years ago
Which order pair is a solution to the system of linear equations -4+y=8 and x-5y=17?
andreev551 [17]

Hello!

To find this ordered pair, we solve for x and y and put them together in an ordered pair.

-4+y=8

x-5y=17

Let's solve the first equation for y.

y= 12

Now, let's plug 12 into the second equation for y.

x-5(12)=17

x-60=17

x=77

Therefore, our ordered pair is (77,12)

I hope this helps!

4 0
3 years ago
Read 2 more answers
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
melamori03 [73]

Answer:

6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs AB=3, BC=4, and AC=5. The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base b and height h is given by A=\frac{1}{2}bh. Therefore, the area of this right triangle is:

A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}

The other triangle is a bit trickier. Triangle \triangle ADC is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

A=\sqrt{s(s-a)(s-b)(s-c)}, where a, b, and c are three sides of the triangle and s is the semi-perimeter (s=\frac{a+b+c}{2}).

The semi-perimeter, s, is:

s=\frac{5+5+4}{2}=\frac{14}{2}=7

Therefore, the area of the isosceles triangle is:

A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}

Thus, the area of the quadrilateral is:

6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}

4 0
3 years ago
Disssss tooo helllllllpppp shoals shzn
iren2701 [21]

Answer:

3/2

Step-by-step explanation:

Use this formula y2-y1/x2-x1

PLUG IN YOUR POINTS

-3 - -9/ 8- 4 =

6/4 (simplify)

1.5 or 3/2

Hope this helps ya!!

7 0
3 years ago
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