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Vladimir [108]
3 years ago
14

Which phrase does not have the same meaning as -3/w

Mathematics
1 answer:
OverLord2011 [107]3 years ago
8 0

Answer:

2 times 2 lol

Step-by-step explanation:

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Getaway Travel Agency surveyed a random sample of 45 of their clients about their vacation plans. Of the clients surveyed, 21 ex
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Choice c trust me i did this
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A triangle has side lengths of 9 in, 13 in, and 20 in. What is the measurement of this triangle's largest angle?
Rama09 [41]

1. Given any triangle ABC with sides BC=a, AC=b and AB=c, the following are true :

     i) the larger the angle, the larger the side in front of it, and the other way around as well. (Sine Law) Let a=20 in, then the largest angle is angle A.

      ii) Given the measures of the sides of a triangle. Then the cosines of any of the angles can be found by the following formula:

       a^{2}=b ^{2}+c ^{2}-2bc(cosA)

2.

20^{2}=9 ^{2}+13 ^{2}-2*9*13(cosA)

400=81+169-234(cosA)   150=-234(cosA)

cosA=150/-234= -0.641

3. m(A) = Arccos(-0.641)≈130°,

4. Remark: We calculate Arccos with a scientific calculator or computer software unless it is one of the well known values, ex Arccos(0.5)=60°, Arccos(-0.5)=120° etc

6 0
4 years ago
Read 2 more answers
Slove and show working for 1/4 ÷ 3
Ronch [10]
.25 divided by 3 is 0.08333333333
6 0
4 years ago
find the equation of the pair of lines perpendicular to the lines pair represented by the equation ax^2-2hxy+by^2=0 and passing
Naddika [18.5K]

The equation of the pair of lines perpendicular to the lines given equation is b x^{2}-2 h x y+a y^{2}=0.

Solution:

Given equation is a x^{2}+2 h x y+b y^{2}=0.

Let m_1 and m_2 be the slopes of the given lines.

Sum of the roots = -\frac{\text {coefficient of } x y}{\text {coefficient of } y^{2}}

               $m_1+m_2=\frac{-2h}{b} – – – – – (1)

Product of the roots = -\frac{\text {coefficient of } x^2}{\text {coefficient of } y^{2}}

                    $m_1 \cdot m_2=\frac{a}{b} – – – – – (2)

The required lines are perpendicular to these lines.

Slopes of the required lines are $-\frac{1}{m_{1}} \text { and }-\frac{1}{m_{2}}

Required lines also passes through the origin,

therefore their y-intercepts are 0.

Hence their equations are:

$y=-\frac{1}{m_{1}}x \text { and }y=-\frac{1}{m_{2}}x

Do cross multiplication, we get

m_1y=-x \  \text{and} \  m_2y=-x

Add x on both sides of the equation, we get

x+m_1y=0 \  \text{and} \  x+m_2y=0

Therefore, the joint equation of the line is

\left(x+m_{1} y\right)\left(x+m_{2} y\right)=0

x^2+m_2xy+m_1xy+m_1m_2y^2=0

x^{2}+\left(m_{1}+m_{2}\right) x y+m_{1} m_{2} y^{2}=0

Substitute (1) and (2), we get

$x^{2}+\left(\frac{-2 h}{b}\right) x y+\left(\frac{a}{b}\right) y^{2}=0

To make the denominator same, multiply and divide first term by b.

$ \frac{b}{b} x^{2}+\left(\frac{-2 h}{b}\right) x y+\left(\frac{a}{b}\right) y^{2}=0

$\frac{bx^2-2hxy+ay^2}{b} = 0

Do cross multiplication, we get

b x^{2}-2 h x y+a y^{2}=0

Hence equation of the pair of lines perpendicular to the lines given equation is b x^{2}-2 h x y+a y^{2}=0.

5 0
4 years ago
5c+16.5=13.5+10c<br> solve for c
lana66690 [7]

Answer:

c = 3/5

Step-by-step explanation:

5c+16.5=13.5+10c

Subtract 5c from both sides:

13.5+5c=16.5

5c=3

c = 3/5

7 0
4 years ago
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