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3 years ago

Help me with this quick pleaseeeee

Mathematics

1 answer:

melisa1 [442]3 years ago

3 0

Answer:

D' = ( 1 , -1 )

Step-by-step explanation:

2 times 1/2 = 1

-2 times 1/2 = -1

hope this helps ^^

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Fuimos al Circo "Hermanos Poncho y Carlos", y tuvimos que pagar en total $350 por mis dos hermanos, mis padres y yo. Mi mamá pag

nalin [4]

Answer:

$160

Step-by-step explanation:

El monto total a pagar es de $350

$\dfrac{1}{7}$ es pagado por mamá

$\dfrac{2}{5}$ es pagado por el hermano mayor

La porción pagada por papá es

$1-(\dfrac{1}{7}+\dfrac{2}{5})=1-(\dfrac{5+14}{35})\\ =1-\dfrac{19}{35}\\ =\dfrac{35-19}{35}\\ =\dfrac{16}{35}$

Esta cantidad pagada por papá es

$350\times\dfrac{16}{35}\\ =10\times 16\\ =\$160$

Esta cantidad pagada por papá es $160.

3 0

3 years ago

Find a vector of magnitude 3 in the direction of v = 10i - 24k. The vector is Oi-Oj+k (Simplify your answer. Use integers or fra

Soloha48 [4]

Given :

A vector v = 10i - 24k.

To Find :

A vector of magnitude 3 in the direction of v = 10i - 24k.

Solution :

Unit vector in the direction of vector v is :

$\^{v}=\dfrac{v}{|v|}\\\\\^{v}=\dfrac{10i-24k}{\sqrt{10^2+24^2}}\\\\\^{v}=\dfrac{10i-24k}{26}$

Now , vector of magnitude 3 in direction v is :

$r=3 \^{v}\\\\r=3\times \dfrac{ (10i-24k)}{26}\\\\r=\dfrac{3}{13} (5i-12k)$

Hence , this is the required solution .

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3 years ago

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faltersainse [42]

47, 57, 67, 77 are the numbers.

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3 years ago

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Ann [662]

Answer:

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Step-by-step explanation:

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3 years ago

A variable force of magnitude F(x) moves a body of mass m along the x-axis from x1 to x2. The net work done by the force in movi

kicyunya [14]

Answer:

$w=\frac{1}{2}m(v^2_2-v^2_1)$

Step-by-step explanation:

We are given that a variable force of magnitude F(x) moves a body along the x- axis from $x_1$ to $x_2$ .

The net work done by the force in moving the body from $x_1$ to $x_2$ .

$v_1$ =Velocity of the body at $x_1$

$v_2=$ Velocity of the body at $x_2$

We know that

Work done=Final kinetic energy-Initial kinetic energy

$w=\frac{1}{2}mv^2_2-\frac{1}{2}mv^2_1$

K.E= $\frac{1}{2}mv^2$

$Net\;work \;done=\frac{1}{2}m(v^2_2-v^2_1)$

Hence, the net work done by the force is given by

$w=\frac{1}{2}m(v^2_2-v^2_1)$

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4 years ago