Calculati cat mai rapid (14+28+36+42):(18+24+46+32)

How you do this ???????

Anon25 [30]

Answer:

the answer is 5 because that is the difference which is what they are asking.

Step-by-step explanation:

12 -7 =5

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8 0

3 years ago

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Find the width of a rectangle with perimeter 92 and length 19

lora16 [44]

Answer:

Step-by-step explanation:

perimeter = l + b + l + b or 2(l + b)

92 = 19 + 19 + 2b

92 = 38 + 2b

2b = 92 - 38

2b = 54

b = 54 / 2

b = 27

breadth or width of the triangle = 27

length = 19

3 0

3 years ago

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Two weeks in a row, the golf course hosts a group of golfers. The second week had 10 more golfers than the first week. Use the d

Goryan [66]

Answer:

A

Step-by-step explanation:

week 1 - 62,68,68,68,69,70,70,72,72,75,75,76,78,78,80,80,80,89,89

week 2 - 62,68,68,68,69,70,70,72,72,72,75,75,76,78,78,80,80,80,85,86,86,86,88,88, 88,88,89,89,89,89

(bold = median)

week 2 median = 79 > week 1 median = 75

5 0

3 years ago

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What is 1+1 im pretty sue ites 5 buut i dot kow

Natasha_Volkova [10]

Answer:

1+1 is 2

Step-by-step explanation:

7 0

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

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

7 0

3 years ago