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3 years ago

WILL GIVE BRAINLIEST

Mathematics

2 answers:

alekssr [168]3 years ago

6 0

Answer: B a parallelogram and then a rectangle

Step-by-step explanation:

GalinKa [24]3 years ago

5 0

Answer:

Step-by-step explanation:

A p e x :3

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Does anyone understand what to do here?

My name is Ann [436]

9514 1404 393

Answer:

see attached

Step-by-step explanation:

There are several possible ways to describe the "type" of a polynomial. Here, since there is a separate column for "degree", we assume that "type" refers to the number of terms.

Polynomials with 1, 2, or 3 terms are called, respectively, monomial, binomial, and trinomial. The first two expressions listed have 1 term only, so are monomials. The last expression has 3 terms, so is a trinomial.

The coefficients are the constant multiplier of the term. Some say a "constant", such as the -8 in the last expression, is not considered a coefficient, because there are no variables that it is multiplying. Here, we have listed it among the coefficients in that expression.

The degree of a term is the sum of the degrees of the variables in the term. For terms with only one variable, it is the exponent of that variable. For terms such as the second expression, the degree is the sum of the exponents: 3+4 = 7. The degree of a polynomial with more than one term is the highest degree of all the terms.

3 0

3 years ago

Julio is paid 1.4 times his normal hourly rate for each hour he works over 30 hours in a week. Last week he worked 35 hours and

Kamila [148]

What is the question?

7 0

3 years ago

A rectangle is to be drawn with perimeter 64cm. If the length is to be 14cm more than the width, determine the are of the rectan

Vikentia [17]

P = 64 cm

2(( x + 14) + x) = 64

2x + 28 + 2x = 64

4x + 28 = 64

4x = 64 - 28

4x = 36

x = 36/4

x = 4

width = 4

lenight = 18

area = l x b

area = 18 x 4

area = 72

5 0

3 years ago

Consider the three points ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 ) . Let ¯ x be the average x-coordinate of these points, and let ¯ y

loris [4]

Answer:

$m=\dfrac{3}{2}$

Step-by-step explanation:

Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

$\overline{x} = \dfrac{x_1+x_2+x_3}{\text{number of points}}$

1) Finding $(\overline{x},\overline{y})$

Average of the x coordinates:

$\overline{x} = \dfrac{1+2+3}{3}$

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

$\overline{y} = \dfrac{3+3+6}{3}$

$\overline{y} = 4$

2) Finding the line through $(\overline{x},\overline{y})$ with slope m.

Given a point and a slope, the equation of a line can be found using:

$(y-y_1)=m(x-x_1)$

in our case this will be

$(y-\overline{y})=m(x-\overline{x})$

$(y-4)=m(x-2)$

$y=mx-2m+4$

this is our equation of the line!

3) Find the squared vertical distances between this line and the three points.

So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.

Distance from point (1,3)

We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?

we'll go back to our equation of the line and use x=1.

$y=m(1)-2m+4$

$y=-m+4$

now we know the two points at x=1: (1,3) and (1,-m+4)

to find the vertical distance we'll subtract the y-coordinates of each point.

$d_1=3-(-m+4)$

$d_1=m-1$

finally, as asked, we'll square the distance

$(d_1)^2=(m-1)^2$

Distance from point (2,3)

we'll do the same as above here:

$y=m(2)-2m+4$

$y=4$

vertical distance between the two points: (2,3) and (2,4)

$d_2=3-4$

$d_2=-1$

squaring:

$(d_2)^2=1$

Distance from point (3,6)

$y=m(3)-2m+4$

$y=m+4$

vertical distance between the two points: (3,6) and (3,m+4)

$d_3=6-(m+4)$

$d_3=2-m$

squaring:

$(d_3)^2=(2-m)^2$

3) Add up all the squared distances, we'll call this value R.

$R=(d_1)^2+(d_2)^2+(d_3)^2$

$R=(m-1)^2+4+(2-m)^2$

4) Find the value of m that makes R minimum.

Looking at the equation above, we can tell that R is a function of m:

$R(m)=(m-1)^2+4+(2-m)^2$

you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)

$\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)$