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3 years ago

Please help.i am really struggling

Mathematics

1 answer:

alina1380 [7]3 years ago

3 0

Answer:

3,-3) becomes ; (3 + 5 , -3-12) ; (8,-15)

(7,-10) becomes;( 7 + 5, -10-12) ; (12,-22)

(13,-14) becomes (7 + 13, -14-10) ; (20,-24)

Step-by-step explanation:

What we have to do here is to add 5 to the x-axis value and subtract 12 from the y-axis value

(3,-3) becomes ; (3 + 5 , -3-12) ; (8,-15)

(7,-10) becomes;( 7 + 5, -10-12) ; (12,-22)

(13,-14) becomes (7 + 13, -14-10) ; (20,-24)

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Answer:

15 / 24
37/52

Step-by-step explanation:

8 .) Probability of students has no pets in senior = No. of outcomes favourable / Total no.of outcome

= 15 / 24

9.) Probability of students has pets in junior = No.of outcomes favourable / Total no. of outcomes

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3 years ago

Quadrilateral ABCD is inscribed in circle O.

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Angle C would equal 59 degrees

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3 years ago

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the diagram shows a 5cm x 5cm x 5cm cube calculate the length of the diagonal AB give your answer correct to 1 decimal place

Lynna [10]

Answer:

√3 * 5 = 5√3 cm

Step-by-step explanation:

→ ABCDEFGH is a cube.

→ CF = Diagonal of cube.

→ CH = Diagonal of Base Face BCDH.

→ Let the side of Each cube = a.

Than,

in Right ∆CFH, By Pythagoras Theoram, we have,

→ CH² + FH² = CF² --------- Equation (1)

and, Similarly, in Right ∆CDH ,

→ CD² + DH² = CH² ------- Equation (2).

Putting Value of Equation (2) in Equation (1), we get,

→ (CD² + DH²) + FH² = CF²

→ a² + a² + a² = CF²

→ CF² = 3a²

→ CF = √3a .

Hence, we can say That Diagonal of a cube is √3 times of its sides.

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Given:-

Side of cube = 5cm.

So,

→ Diagonal of cube = √3 * 5 = 5√3 cm. (Ans.)

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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.

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3 years ago

Lily buys a sweater that is on sale for $42, which is 75% of the original price of the sweater. What was the original price of t

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So first sivid 42 by 3 to get 14. Then multiply 14 by 4 to get 56. So the answer is C

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3 years ago

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