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3 years ago

TRUE

Mathematics

1 answer:

Kipish [7]3 years ago

5 0

run me my points .

Step-by-step explanation:

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What is -2x+5>11 graphed out on a number line?

grin007 [14]

Step-by-step explanation:

-2x>11-5

x<6/-2

x<-3

Make sure the circle in the number line isnt coloured in!

5 0

3 years ago

(50 POINTS AND BRAINLIEST) Show all work to multiply quantity 2 plus the square root of negative 25 end quantity times quantity

xeze [42]

The equation would be (3+√-16)(6-√-64)
A negative square root can not be a real number, meaning that we have to use the imaginary number.
(3+4i)(6-8i)
How to multiply imaginary numbers...
Treat this like an equation looking like (x-a)(y-b)
Use FOIL
(3+4i)(6-8i)
= 18-24i+24i-32i²
= 18-32i² (i²= -1 because i= √-1)
= 18- (-32)
= 18 + 32
= 50

8 0

3 years ago

Read 2 more answers

Felice knows that segment RS || segment QT. She wants to use the definition of a parallelogram to prove that the quadrilateral i

tangare [24]

Answer:

see below

Step-by-step explanation:

The definition of a parallelogram is that opposite sides of the quadrilateral are parallel. Felice already knows one pair of opposite sides is parallel. By showing the slope of RQ is the same as the slope of ST, she can show the other pair of opposite sides is parallel, hence the figure is a parallelogram.

The other answer choices are essentially nonsense.

8 0

3 years ago

17.2 writes it as a mixed number in lowest terms

bixtya [17]

1/5 as a mixed number

3 0

3 years ago

1 pt) If a parametric surface given by r1(u,v)=f(u,v)i+g(u,v)j+h(u,v)k and −4≤u≤4,−4≤v≤4, has surface area equal to 1, what is t

natta225 [31]

The area of the surface given by $\vec r_1(u,v)$ is 1. In terms of a surface integral, we have

$1=\displaystyle\int_{-4}^4\int_{-4}^4\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

By multiplying each component in $\vec r_1$ by 5, we have

$\dfrac{\partial\vec r_2(u,v)}{\partial u}=5\dfrac{\partial\vec r_1(u,v)}{\partial u}$

and the same goes for the derivative with respect to $v$ . Then the area of the surface given by $\vec r_2(u,v)$ is

$\displaystyle\int_{-4}^4\int_{-4}^425\left\|\frac{\partial\vec r_1(u,v)}{\partial u}\times\frac{\partial\vec r_1(u,v)}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\boxed{25}$

8 0

3 years ago