What did Zorna say about marrying a shorter man?

Mathematics

1 answer:

Aloiza [94]3 years ago

5 0

"Better to have loved a short man than never to have loved a tall".

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Is line 1 (1,5) (3,-2) and like 2 (-3,2) (4,0) perpendicular of parallel?

Lera25 [3.4K]

Slope Formula: $\frac{y_2-y_1}{x_2-x_1}$

So remember that <u>perpendicular lines have slopes that are negative reciprocals to each other</u> and <u>parallel lines have the same slope.</u> To find out if they are either parallel or perpendicular, plug the pair of points into the slope formula to find their slopes:

$\textsf{Line 1}\\\\\frac{5-(-2)}{1-3}=-\frac{7}{2}\\\\\textsf{Line 2}\\\\\frac{0-2}{4-(-3)}=-\frac{2}{7}$

Since these slopes aren't the same nor are they negative reciprocals to each other, <u>the lines are neither parallel nor perpendicular.</u>

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3 years ago

Please help I need to know the answer

Kazeer [188]

Answer:

Original price: 51$
Discount percentage: 40%

-----------

Details
Discount = Original Price x Discount %/100
Discount = 51 × 40/100
Discount = 51 x 0.4
You save = $20.40

Final Price = Original Price - Discount
Final Price = 51 - 20.4
Final Price = $30.60

there you go, coupon A is 20.40$ Dollars less than the price of coupon B, your welcome :)) i wish u best of luck in the exam:)

also, please mark me as brainliest, it would really help me out, thanks :))

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2 years ago

If m =2 solve these 3m²-2m-7

almond37 [142]

Answer:

Step-by-step explanation:

m=2

3m²- 2m- 7

3(2)² -2(2) -7

3(4)- 4- 7

12- 4- 7= 1

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3 years ago

Solve for y.<br> -38 = 7v+17 – 12v

iren2701 [21]

Answer:

Step-by-step explanation:

$-38=7v+17-12v\qquad\text{subtract 17 from both sides}\\\\-38-17=7v+17-17-12v\qquad\text{combine like terms}\\\\-55=-5v\qquad\text{divide both sides by (-5)}\\\\\dfrac{-55}{-5}=\dfrac{-5v}{-5}\\\\11=v\to v=11$

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3 years ago

E=mc2 is one of the most recognizable equations in the world. In this task, you’ll attempt to understand it better. Let’s start

Mashutka [201]

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
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E=mc^2\implies \cfrac{E}{m}=c^2\implies \sqrt{\cfrac{E}{m}}=c\implies \left( \frac{E}{m} \right)^{\frac{1}{2}}=c$

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3 years ago