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3 years ago

Hi doesanyone know answer

Mathematics

2 answers:

Lilit [14]3 years ago

8 0

Answer:

170

Step-by-step explanation:

Mashutka [201]3 years ago

3 0

Answer:

170 is the answer good luck

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3 years ago

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All 4 questions please

scZoUnD [109]

Answer:

π = S/2πrh
v = √2E/m
b = (-7a+c)/2
x =3y-1

Step-by-step explanation:

$S = 2\pi rh\\Divide\:both\sides\:of\:the\:equation\:by\:2rh\\\\\frac{S}{2rh} = \frac{2 \pi rh}{2 rh} \\\\\pi = \frac{S}{2rh}$

$E = \frac{1}{2} mv^2\\\\E = \frac{mv^2}{2} \\\\Cross\:multiply\\\\2E =mv^2\\Divide\:both\:sides\:of\:the\:equation\:by\:m\\\frac{2E}{m} = \frac{mv^2}{m} \\\\\frac{2E}{m} =v^2\\Square\:root\:both\:sides\:of\:the\:equation\\\sqrt{\frac{2E}{m} } = \sqrt{v^2} \\\\\sqrt{\frac{2E}{m} } =v\\\\v =\sqrt{\frac{2E}{m} }$

$7a+2b =c\\Isolate\:b\\\\2b =c-7a\\2b= -7a+c\\Divide\:both\:sides\:of\:the\:equation\:by \:2\\\frac{2b}{2} =\frac{-7a+c}{2} \\\\b = \frac{-7a+c}{2}$

$y = \frac{1}{3} (x+1)\\\\y = \frac{x+1}{3}\\\\Cross\:multiply \\3y =x+1\\Isolate \:x\\3y -1=x\\\\x =3y-1$

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3 years ago

Help I need it done now

Paha777 [63]

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3 years ago

The watch at Macys is $400. The store is having a sale and everything is 30% off. What is the discount amount?

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3 years ago

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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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3 years ago