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2 years ago

Can a triangle be made with sides of length 7, 10 and 20?

Mathematics

1 answer:

salantis [7]2 years ago

3 0

Yes it can its called a scalene triangle

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I. Direction: Write the following statement as equations. Write your answers

Bingel [31]

Answer:

1. 4x - 29=11

2. (3x - 4) (2) = 46

3. 3/5x + 8 = 50

4. The difference of 51 and the sum of x and 5.

5. The product of 3 and y minus eight.

Step-by-step explanation:

4 0

2 years ago

PLZ HELP ASAP DIVIDE LINE SEGMENTS

rodikova [14]

Segment in the direction from A to C
Initial Point: A=(9,5)=(xa,ya)→xi=xa=9, yi=ya=5
Final point: C=(-7,1)=(xc,yc)→xf=xc=-7, yf=yc=1
B=(xb,yb)=?
Proportion: r=AB/BC=3:1=3/1→r=3

xb=(xi+r*xf)/(1+r)
Replacing xi=xa=9, xf=xc=-7 and r=3
xb=[9+3*(-7)]/(1+3)
xb=(9-21)/4
xb=(-12)/4
xb=-3

yb=(yi+r*yf)/(1+r)
Replacing yi=ya=5, yf=yc=1 and r=3
yb=[5+3*(1)]/(1+3)
yb=(5+3)/4
yb=8/4
yb=2

B=(xb,yb)→B=(-3,2)

Answer: B=(-3,2)

7 0

2 years ago

Simplify the expression: -6(1.4 - r)=

Vitek1552 [10]

Answer:

$6r-8.4$

Step-by-step explanation:

$-6(1.4-r)$

Use the distibutivve property to multiply -6 by 1.4 - r

$-8.4 +6r$

Then move over 6r to the front

$6r-8.4$

6 0

2 years ago

Read 2 more answers

Determine which of the sets of vectors is linearly independent. A: The set where p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t B: The se

defon

Answer:

The set of vectors A and C are linearly independent.

Step-by-step explanation:

A set of vector is linearly independent if and only if the linear combination of these vector can only be equalised to zero only if all coefficients are zeroes. Let is evaluate each set algraically:

$p_{1}(t) = 1$ , $p_{2}(t)= t^{2}$ and $p_{3}(t) = 3 + 3\cdot t$ :

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (3 +3\cdot t) = 0$

$(\alpha_{1}+3\cdot \alpha_{3})\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot t = 0$

The following system of linear equations is obtained:

$\alpha_{1} + 3\cdot \alpha_{3} = 0$

$\alpha_{2} = 0$

$\alpha_{3} = 0$

Whose solution is $\alpha_{1} = \alpha_{2} = \alpha_{3} = 0$ , which means that the set of vectors is linearly independent.

$p_{1}(t) = t$ , $p_{2}(t) = t^{2}$ and $p_{3}(t) = 2\cdot t + 3\cdot t^{2}$

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot t + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (2\cdot t + 3\cdot t^{2})=0$

$(\alpha_{1}+2\cdot \alpha_{3})\cdot t + (\alpha_{2}+3\cdot \alpha_{3})\cdot t^{2} = 0$

The following system of linear equations is obtained:

$\alpha_{1}+2\cdot \alpha_{3} = 0$

$\alpha_{2}+3\cdot \alpha_{3} = 0$

Since the number of variables is greater than the number of equations, let suppose that $\alpha_{3} = k$ , where $k\in\mathbb{R}$ . Then, the following relationships are consequently found: