All categories

3 years ago

Determine the product

Mathematics

2 answers:

UkoKoshka [18]3 years ago

7 0

The answer is -6xy+6xz-10x^(2)

astraxan [27]3 years ago

6 0

= -10x^2 - 6xy + 6xz

hope it helps

You might be interested in

Can fractions still be considered natural, whole, integers, rational, irrational, or real numbers?

vitfil [10]

<span>Fractions are numbers which can be represented in the form p/q, where p and q are WHOLE NUMBERS and q is not zero Rational numbers are numbers which can be represented in the form p/q where p and q are INTEGERS and q is not zero So your saying " all rational numbers are all numbers that can be represented as a fraction." is wrong....... all fractions are rational numbers but NOT ALL rational numbers are fractions..</span>

7 0

2 years ago

Use technology to approximate the solution(s) to the system of equations to the nearest tenth of a unit.

kherson [118]

Answer:

A.) (3.6, 0.6)
D.) (2.6, 0.4)

Step-by-step explanation:

See below for a graph.

___

Choices B, C, E can be eliminated on the basis that neither x nor g(x) can be negative. The domain of f(x) is x>0; the range of g(x) is x≥0.

7 0

2 years ago

4^x+7 = 8^2x-3 Solve each equation showing work please

RUDIKE [14]

If 4=2^2

8=2^3

(2^)2x+10=(2^)6x

(2^)6x-(2^)2x=10

(2^)2x×7=10, where x has no real value

3 0

3 years ago

Need help with full work quick

vekshin1

Answer:

Step-by-step explanation:

36x^3 - 81x

9x(4x^2 - 9)

9x(2x - 3)(2x + 3)

She is right

4 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