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3 years ago

Use the distributive property to write an expres sion that is equivalent to-2(4x-5)

Mathematics

1 answer:

Novosadov [1.4K]3 years ago

5 0

Answer:

-8x+10

Step-by-step explanation:

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4 Points

Vladimir [108]

Answer:

$35$

Step-by-step explanation:

One is given the following function ( $y=3x+5$ ). The given problem asks one to find the output of the function when the input is (10). Keep in mind, a general rule for dealing with functions is that, when present, the variable (x) represents the input. Whereas the variable (y) represents the output. Thus, if one were to substitute (10) in place of (x), and simplify: the result one would get is the output. Apply this idea to the given scenario:

$y=3x+5$

$y=3(10)+5$

$y=30+5$

$y=35$

4 0

2 years ago

Anyone knows how to do integration? Pls help! URGENT!

MrRa [10]

Uhhhhhhh I don't really know look on quizlet

8 0

2 years ago

A set of equations is given below.

spayn [35]

Answer:

Step-by-step explanation:

lucky guess from my boi tony

7 0

2 years ago

Z+x3:use x = 1 and z = 19

pogonyaev

Answer:21

Step-by-step explanation: plug in 19 and 1

1x3 is 3 plus 19 is 21

7 0

2 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

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

7 0

3 years ago