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goldenfox [79]
3 years ago
5

If DE = 37 cm and EF = 16 cm, then what are the possible lengths for DF so that DE,EF, and DF can form a triangle? Explain your

reasoning.
Mathematics
1 answer:
chubhunter [2.5K]3 years ago
4 0

Answer:

The length of DF must be between 21 and 53.

Step-by-step explanation:

In a triangle, the length of two sides added together must exceed the length of the 3rd side. So, since EF is the shortest of the two givens, we know that EF + DF must be greater than DE. So we can plug in these numbers to find the minimum.

EF + DF > DE

16 + DF > 37

DF > 21

Now, for the upper maximum, we know that the two given lengths must be greater than the length of DF. So again, we can solve for the maximum using the amounts.

DE + EF > DF

37 + 16 > DF

53 > DF

With these two in mind, we know that DF must be between 21 and 53

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6 0
2 years ago
1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t
natta225 [31]

The area of the surface given by \vec r_1(u,v) is 1. In terms of a surface integral, we have

1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv

By multiplying each component in \vec r_1 by 5, we have

\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}

and the same goes for the derivative with respect to v. Then the area of the surface given by \vec r_2(u,v) is

\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}

8 0
3 years ago
Simplify 27^2.<br><br> please (thxx)
adoni [48]

Answer:

729

Step-by-step explanation:

Raise 27 to the power of 2.

729

4 0
3 years ago
A.
Reptile [31]

Answer:

The real zeros of f(x) are x = 0.3 and x = -3.3.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this problem, we have that:

f(x) = x^{2} + 3x - 1

So

a = 1, b = 3, c = -1

\bigtriangleup = 3^{2} - 4*1*(-1) = 13

x_{1} = \frac{-3 + \sqrt{13}}{2*1} = 0.3

x_{2} = \frac{-3 - \sqrt{13}}{2*1} = -3.3

The real zeros of f(x) are x = 0.3 and x = -3.3.

7 0
3 years ago
A clerk is marking up merchandise 34%. The original price of an item is $455. What will be the retail price?
Nadusha1986 [10]
To find markup price, multiply the original number (455) by 1 + the decimal equivilent of the percent increase (.34)
so
455*1.34= $609.70
4 0
2 years ago
Read 2 more answers
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