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3 years ago

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Which graph shows the system StartLayout Enlarged left-brace 1st row x squared + y = 2 2nd row x squared + y squared = 9 EndLayo

ut?

1 answer:

zimovet [89]3 years ago

7 0

Answer: Hope this helps <3

Step-by-step explanation:

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Simplify the expression 7.8n²+0.5+9n-.0.4-7n+n²

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The final answer for this question would be.....

8.8n^2+16n+0.1

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For every 4 hours Michelle works, she gets a 15 minute break. Which equation shows the proportional relationship between the num

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b=8w

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Romashka-Z-Leto [24]

Answer:

The interquartile range is <em>50.</em>

Step-by-step explanation:

To find our answer we have to first <em>quartile 1</em> and <em>quartile 3</em> are equal too. When we look at the plot <em>quartile 1 </em>is equal to <em>20,</em> <em>quartile 3 </em>is equal to <em>70</em> because it is in between <em>60</em> and <em>80</em>. Now to find the interquartile range we will <em>subtract 70</em> from <em>20</em> and we get <em>50</em>. Therefore, <u><em>50</em></u><em> is our answer.</em>

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Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl

GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane $n=(-2,-2,1)$ , then r will have the next parametric equations:

$x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda$

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

$-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3$

Substitute the value of $\lambda$ in the parametric equations:

$x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\$

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

$d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u$

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What is the difference to the nearest hundredth of a unit between the two zeros

Luda [366]

Answer:

0.1

Step-by-step explanation:

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