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Juli2301 [7.4K]

3 years ago

10

Which of the following demonstrates the Commutative Property of Multiplication?

2 answers:

vova2212 [387]3 years ago

6 0

The answer is B3 parentheses four lowercase

Likurg_2 [28]3 years ago

3 0

I believe the answer is d.

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What number do you multiply to find this partial product 8×56

Serhud [2]

Answer: It is 7

Step-by-step explanation:

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2 years ago

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A grocery store has 120 bottles of spring

notka56 [123]

Question:

A grocery store has 120 bottles of spring water in stock. The store orders bottles of spring water in cases of 24. The store wants to order enough cases of spring water so it has more than 500 bottles in stock. Which inequality best models this situation?

A. 24x + 120 > 500

B. 24x - 120 > 500

C. 24x + 500 > 120

D. 24(x+120) > 500

Answer:

Option A. 24x + 120 > 500 is the inequality that best models this situation.

Step-by-step explanation:

The Number of bottles of spring water in the stock = 120

Each case of the spring water bottles contains 24 bottles

The store wants to order enough cases of spring water so that it has over 500 bottles in stock.

So there are already 120 bottles in stock .

The store will order

24x +120 > 500

where x is the number of spring water bottle cases

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3 years ago

A model airplane flew a distance of 330 feet in 15 seconds.

kondaur [170]

So whats the question

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3 years ago

WILL GIVE BRAINLIEST ANSWER!

DENIUS [597]

6. No, the conclusions wrong because 20 out of 50 peoples favourite is drawing which means 20/50 which equals 0.4 times 100 which is 40% not 60%

7. Yes, the conclusion is right because 5/50 is equal to 0.1 times 100 which is 10%

4 0

3 years ago

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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