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What is 5/12 times 6 in multiply fractions with whole numbers

KiRa [710]

5/12*6 =5/12*6/1 crosscancle you get 5/2 the you get 2 1/2

3 0

4 years ago

a mother-of-four nose dolphin tale about 278 pounds of food in one week about how much food does he eat in a day

BigorU [14]

About 39 pounds each day of the week

7 0

4 years ago

Read 2 more answers

If neither a nor b are equal to zero, which answer most accurately describes the product of (a + bi)(a - bi)?

Scrat [10]

Answer: the correct option is

(D) The imaginary part is zero.

Step-by-step explanation: Given that neither a nor b are equal to zero.

We are to select the correct statement that accurately describes the following product :

$P=(a+bi)(a-bi)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We will be using the following formula :

$(x+y)(x-y)=x^2-y^2.$

From product (i), we get

$P\\\\=(a+bi)(a-bi)\\\\=a^2-(bi)^2\\\\=a^2-b^2i^2\\\\=a^2-b^2\times (-1)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }i^2=-1]\\\\=a^2+b^2.$

So, there is no imaginary part in the given product.

Thus, the correct option is

(D) The imaginary part is zero.

3 0

3 years ago

I REALLY NEED HELP please please help me answer all please

Radda [10]

9514 1404 393

Answer:

all are functions except Relation 2

Step-by-step explanation:

If any input value (first of an ordered pair) is used more than once, the relation is not a function. Output values can be used as often as you like.

Only Relation 2 is not a function.

3 0

3 years ago

Find the orthocenter for the triangle described by each set of vertices.

givi [52]

Answer:

The orthocentre of the given vertices ( 2 , -3.5)

Step-by-step explanation:

Step(i):-

The orthocentre is the intersecting point for all the altitudes of the triangle.

The point where the altitudes of a triangle meet is known as the orthocentre.

Given Points are K (3.-3), L (2,1), M (4,-3)

The Altitudes are perpendicular line from one side of the triangle to the opposite vertex

The altitudes are MN , KO , LP

step(ii):-

Slope of the line

$KL = \frac{y_{2}-y_{1} }{x_{2}-x_{1} } = \frac{1-(-3)}{2-3} = -4$

The slope of MN =

The perpendicular slope of KL

= $\frac{-1}{m} = \frac{-1}{-4} = \frac{1}{4}$

The equation of the altitude

$y - y_{1} = m( x-x_{1} )$

$y - (-3) = \frac{1}{4} ( x-4 )$

4y +12 = x -4

x - 4 y -16 = 0 ...(i)

Step(iii):-

Slope of the line

$LM = \frac{y_{2}-y_{1} }{x_{2}-x_{1} } = \frac{-3-1}{4-2} = -2$

The slope of KO =

The perpendicular slope of LM

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

The equation of the altitude

$y - y_{1} = m( x-x_{1} )$

The equation of the line passing through the point K ( 3,-3) and slope

m = 1/2

$y - (-3) = \frac{1}{2} ( x-3 )$

2y +6 = x -3

x - 2y -9 =0 ....(ii)

Solving equation (i) and (ii) , we get

subtracting equation (i) and (ii) , we get

x - 4y -16 -( x-2y-9) =0

- 2y -7 =0

-2y = 7

y = - 3.5

Substitute y = -3.5 in equation x -4y-16=0

x - 4( -3.5) - 16 =0

x +14-16 =0

x -2 =0

x = 2

The orthocentre of the given vertices ( 2 , -3.5)

4 0

3 years ago