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3 years ago

Which measure is equivalent to 1.5 kg?

Mathematics

2 answers:

DanielleElmas [232]3 years ago

8 0

Answer:

1500 grams.

Step-by-step explanation:

1.5 kg = 1500 grams

diamong [38]3 years ago

3 0

Answer:

1500 grams.

Step-by-step explanation:

1500 grams.

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In the given figure, find the value of a and b.

Sladkaya [172]

Answer is 180- (60+70) = 50°

6 0

2 years ago

Read 2 more answers

Need help Thank you very much

Brrunno [24]

Answer:

A is right I believe

Step-by-step explanation:

a: 5x

b: 2x

c: 1/5x

d: 1/2x

4 0

3 years ago

Respuesta de (-0.5)²

Vladimir79 [104]

Answer:

0.25

Step-by-step explanation:

we have

$(-0.5)^2$

we know that

$-0.5=-\frac{1}{2}$

substitute

$(-\frac{1}{2})^2=(-\frac{1}{2})(-\frac{1}{2})$

Remember that

$(-1)(-1)=1\\(-1)(1)=(-1)$

$(-\frac{1}{2})(-\frac{1}{2})=\frac{1}{4}=0.25$

5 0

3 years ago

Why the output is a zero number ?<br><img src="https://tex.z-dn.net/?f=%28x%20-%203%29%20%5Ctimes%20%28x%20%20%20%2B%204%29%20%3

ELEN [110]

Answer:

see explanation

Step-by-step explanation:

Given

(x - 3) × (x + 4) = 0, that is

(x - 3)(x + 4) = 0

The zero product indicates that (x - 3) = 0 or (x + 4) = 0

x - 3 = 0 ⇒ x = 3

x + 4 = 0 ⇒ x = - 4

Thus if x = 3, then

(3 - 3)(x + 4) = 0 × (x + 4) = 0

Similarly if x = - 4 the output is zero

4 0

3 years ago

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4

dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

$x_1\ =\ -2$

$y_1\ =\ 3$

$z_1\ =\ -4$

Distance is measure across the line

$\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}$

So, we can write

$\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k$

$=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k$

$=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k$

$=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}$

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

$3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3$

$=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18$

$=>18k-24+12k+9+10k-32\ =\ 18$

$=>\ k\ =\dfrac{13}{8}$

So,

$x\ =\ 3k-4$

$=\ 3\times \dfrac{13}{8}-4$

$=\ \dfrac{7}{4}$

$y\ =\ \dfrac{4k+3}{2}$

$=\ \dfrac{4\times \dfrac{13}{8}+3}{2}$

$=\ \dfrac{19}{4}$

$z\ =\ \dfrac{5k-16}{3}$

$=\ \dfrac{5\times \dfrac{13}{8}-16}{3}$

$=\ \dfrac{-21}{8}$

Now, the distance of point from the plane is given by,

$d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$

$=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}$

$=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}$

$=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}$

$=\ \sqrt{\dfrac{1177}{64}}$

$=\ 4.28$

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

= 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0

3 years ago