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3 years ago

PLEASE HELP I WILL GIVE BRAINLIEST TO RIGHT ANSWERS ONLY

Mathematics

1 answer:

Fofino [41]3 years ago

4 0

Answer:

logb(x y) = y ∙ logb(x)

Step-by-step explanation:

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A clothing store sells men's hats for $30 each. At this price, the store sells 100 men's hats per week. The owner estimates that

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Answer:

Step-by-step explanation:

At this price, the store sells 100 men's hats per week. The owner estimates that for every $1 increase in price, one fewer men's hat is sold per week

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2 years ago

Rachel weighed her two cats, Tiger and Mittens. Tiger weighed 10.91 pounds, and Mittens weighed 178.08 ounces. Which cat is heav

Dafna11 [192]

Answer:

answer is c

Step-by-step explanation:

178.08 oz = 11.13 lb

11.13-10.91 = 0.22 lb

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2 years ago

Given the geometric sequence where a1 = 3 and r = √2 find a9

Zigmanuir [339]

Answer:

$a_9=48$

Step-by-step explanation:

we are given

sequence is geometric

so, we can use nth term formula

$a_n=a_1(r)^{n-1}$

we have

$a_1=3$

$r=\sqrt{2}$

we have to find a9

so, we can plug n=9

we get

$a_9=3(\sqrt{2})^{9-1}$

$a_9=2^4\cdot \:3$

$a_9=48$

8 0

3 years ago

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

What is the surface area of the prism?

IRINA_888 [86]

Its b hope this helps

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3 years ago

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