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3 years ago

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Jasmine biked 33 miles last week. Deonte biked 28 miles last week. If Deonte biked the same number of miles the four days he bik

ed last week, how many miles did he bike each day?

1 answer:

Andrej [43]3 years ago

5 0

28/4=7 so Deonte biked 7 miles each day

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C+s=48
2c+4s=134 ⇒ c+2s=67

subtracting the equations, we get, s=67-48=19,
so, c=48-19=29

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3 years ago

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Choose the expression that represents a cubic expression.

Anna007 [38]

Answer:

10x³ - 6x² - 9x + 12 is a cubic expression ⇒ 2nd

6x + 6 is a linear expression ⇒ 1st

5x² + 9x - 1 is a quadratic expression ⇒ 2nd

Quadratic expression ⇒ 1st

Step-by-step explanation:

* Lets explain the degree of any expression

- The degree of any expression is the greatest power of the variable

- Ex:

# 3x - 3 is a first degree expression

# x² - 3x + 2 is a second degree expression

# 5x³ - 2x² + 5x - 2 is a third degree expression

- The first degree expression is called linear expression

- The second degree expression is called quadratic expression

- The third degree expression is called cubic expression

* Lets solve the problem

∵ The cubic expression is the expression with greatest power of 3

∴ 10x³ - 6x² - 9x + 12 is the cubic expression

∵ The linear expression is the expression with greatest power of 1

∴ 6x + 6 is a linear expression

∵ The quadratic expression is the expression with greatest power of 2

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∵ The expression is -2x² + 11

∵ The greatest power of x is 2

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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.

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Well I believe it's b

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Answer:

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Step-by-step explanation:

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3 years ago