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3 years ago

The coordinates of the vertices of quadrilateral JKLM are J(-3,2), K(4,-1), L(2,-5) and M(-5,-2).

1 answer:

5 0

Slope of JK=((-1)-2)/(4-(-3)=-3/7 Slope of KL=((-5)-(-1)/(2-4=2 Slope of LM=((-2)-(-5))/(-5-2)=-3/7 Slope of MJ=(2-(-2))/((-3)-(-5))= 2 JK is parallel to LM and KL is parallel to MJ. Therefore JKLM is a parallelogram.

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Separate the number 41 and the two parts so that the first number is eight more than twice the second number what are the two nu

vesna_86 [32]

The two numbers are 30 and 11

Solution:

Given that we have to separate the number 41 into two parts

Let the second number be "x"

Given that first number is eight more than twice the second number

first number = eight more than twice the second number

first number = 8 + twice the "x"

first number = 8 + 2x

So we can say first number added with second number ends up in 41

first number + second number = 41

8 + 2x + x = 41

8 + 3x = 41

3x = 41 - 8

3x = 33

x = 11

first number = 8 + 2x = 8 + 2(11) = 8 + 22 = 30

Thus the two numbers are 30 and 11

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3 years ago

Write the slope-intercept form of the equation that passes through the point (-3, 5) and is perpendicular to the line y = 1/5x +

aleksley [76]

Answer:

Step-by-step explanation:

$\text{Let}\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\=========================$

$\text{We have}\ y=\dfrac{1}{5}x+10\to m_1=\dfrac{1}{5}\\\\\text{Therefore}\ m_2=-\dfrac{1}{\frac{1}{5}}=-5.\\\\\text{Put the value of a slope and the coordinates of the point (-3, 5)}\\\text{to the equation}\ y=mx+b:\\\\5=-5(-3)+b\\5=15+b\qquad\text{subtract 15 from both sides}\\-10=b\to b=-10\\\\\text{Finally:}\\\\y=-5x-10$

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3 years ago

My phone rang exactly midway between 10:53 AM one day and 1:05 PM the next day. What time did it ring?

marishachu [46]

The time it rang was 11:59 PM.

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3 years ago

Read 2 more answers

Sin α = 21/29, α lies in quadrant II, and cos β = 15/17, β lies in quadrant I Find sin (α - β).

Sever21 [200]

$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$

$\sin\alpha=\dfrac{21}{29}\implies \cos^2\alpha=1-\sin^2\alpha=\dfrac{400}{841}$

Since $\alpha$ lies in quadrant II, we have $\cos\alpha$ , so

$\cos\alpha=-\sqrt{\dfrac{400}{841}}=-\dfrac{20}{29}$

$\cos\beta=\dfrac{15}{17}\implies\sin^2\beta=1-\cos^2\beta=\dfrac{64}{289}$

$\beta$ lies in quadrant I, so $\sin\beta>0$ and

$\sin\beta=\sqrt{\dfrac{64}{289}}=\dfrac8{17}$

So

$\sin(\alpha-\beta)=\dfrac{21}{29}\dfrac{15}{17}-\left(-\dfrac{20}{29}\right)\dfrac8{17}=\dfrac{475}{493}$

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son4ous [18]

Answer:

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