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3 years ago

A triangle has two side lengths of 7 and 12. what value could the length of the third side be?

Mathematics

2 answers:

Zolol [24]3 years ago

8 0

7x12 =52 so that is your answer 52

erik [133]3 years ago

6 0

Answer:

13.9

Step-by-step explanation:

7^2 + 12^2 = c^2

49 + 144 = 193

square root 193

and it’s 13.9

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3 years ago

The height in meters of a toy rocket at t seconds can be found by the

OlgaM077 [116]

Answer: 81.2

Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Using the pattern, give the coefficients of (x + y)^5 and (x + y)^6

musickatia [10]

Answer:

$(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$

Step-by-step explanation:

In order to find the values of $(x+y)^5$ and $(x+y)^6$ , you need to apply the binomial theorem (high-level math you most likely don't need to worry about, it's easier than multiplying all the binomials together).

$(x+y)^5$ = $\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$ = $\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$ = $x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$ .

$(x+y)^6$ = $\sum _{i=0}^6\binom{6}{i}x^{\left(6-i\right)}y^i$ = $\frac{6!}{0!\left(6-0\right)!}x^6y^0+\frac{6!}{1!\left(6-1\right)!}x^5y^1+\frac{6!}{2!\left(6-2\right)!}x^4y^2+\frac{6!}{3!\left(6-3\right)!}x^3y^3+\frac{6!}{4!\left(6-4\right)!}x^2y^4+\frac{6!}{5!\left(6-5\right)!}x^1y^5+\frac{6!}{6!\left(6-6\right)!}x^0y^6$ = $x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$ .

5 0

3 years ago

A pyramid with base area of 16sq cm and height of 30cm

motikmotik

b*h/3 is the formula to find the volume (or area whichever it is) of a pyramid so it will be 16 times 30 divided by 3 which would get u to 480/3 which would then get u to 160 cm

3 0

3 years ago