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1 year ago

12x+5y3z, where x=12, y = 6, and z = 3

Mathematics

1 answer:

mestny [16]1 year ago

8 0

Answer:

12×x+5y×3z

12×12+5×6×3×3

144+30×9

174×9

1,566.

that's what I found out.

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What is a correct congruence statement for the triangles shown?

Kamila [148]

Answer:

ΔRHS ≅ ΔNKW

Step-by-step explanation:

We are given that, in the triangles RHS and NKW.

∠H = ∠K

Side HS = KW.

∠S = ∠W

Now, using the ASA congruence criteria, which states that when two corresponding angles of the triangles are equal with the included side between them is equal, then both the triangles are congruent.

Hence, according to ASA criteria, ΔRHS ≅ ΔNKW.

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2 years ago

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How can you use a drawing to construct an argument? (ONLY IF YOU KNOW)

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Draw something personally offensive to the viewer (nothing inappropriate). Or you could simply add the subjects of your argument into the drawing.

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2 years ago

1. S(–4, –4), P(4, –2), A(6, 6) and Z(–2, 4) a) Apply the distance formula for each side to determine whether SPAZ is equilatera

Aleksandr [31]

Answer:

a) SPAZ is equilateral.

b) Diagonals SA and PZ are perpendicular to each other.

c) Diagonals SA and PZ bisect each other.

Step-by-step explanation:

At first we form the triangle with the help of a graphing tool and whose result is attached below. It seems to be a paralellogram.

a) If figure is equilateral, then SP = PA = AZ = ZS:

$SP = \sqrt{[4-(-4)]^{2}+[(-2)-(-4)]^{2}}$

$SP \approx 8.246$

$PA = \sqrt{(6-4)^{2}+[6-(-2)]^{2}}$

$PA \approx 8.246$

$AZ =\sqrt{(-2-6)^{2}+(4-6)^{2}}$

$AZ \approx 8.246$

$ZS = \sqrt{[-4-(-2)]^{2}+(-4-4)^{2}}$

$ZS \approx 8.246$

Therefore, SPAZ is equilateral.

b) We use the slope formula to determine the inclination of diagonals SA and PZ:

$m_{SA} = \frac{6-(-4)}{6-(-4)}$

$m_{SA} = 1$

$m_{PZ} = \frac{4-(-2)}{-2-4}$

$m_{PZ} = -1$

Since $m_{SA}\cdot m_{PZ} = -1$ , diagonals SA and PZ are perpendicular to each other.

c) The diagonals bisect each other if and only if both have the same midpoint. Now we proceed to determine the midpoints of each diagonal:

$M_{SA} = \frac{1}{2}\cdot S(x,y) + \frac{1}{2}\cdot A(x,y)$

$M_{SA} = \frac{1}{2}\cdot (-4,-4)+\frac{1}{2}\cdot (6,6)$

$M_{SA} = (-2,-2)+(3,3)$

$M_{SA} = (1,1)$

$M_{PZ} = \frac{1}{2}\cdot P(x,y) + \frac{1}{2}\cdot Z(x,y)$

$M_{PZ} = \frac{1}{2}\cdot (4,-2)+\frac{1}{2}\cdot (-2,4)$

$M_{PZ} = (2,-1)+(-1,2)$

$M_{PZ} = (1,1)$

Then, the diagonals SA and PZ bisect each other.

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2 years ago

Directions:Use the product rule to simplify the following monomials.

Ann [662]

Answer:

See answers below

Step-by-step explanation:

6) -12(a^5)(b^6)

7) -24(x^8)(y^11)

8) -30y^6

9) -6(x^4)(y^5)

10) 42(m^3)(n^5)

Have a great day :)

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2 years ago

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Helpppppp huwdhdidcvqicgqe8ygc

lana [24]

YESSS USE THATTTTTTTTTTTT

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2 years ago

12x+5y3z, where ​x=12​, y = 6, and z = 3

12x+5y3z, where x=12, y = 6, and z = 3