For each of the following quadrilaterals, Select all the properties that must be true.

How can you tell if a function is linear?

neonofarm [45]

Linear functions are usually of degree 1. They are a straight line. y=mx+b

3 0

3 years ago

Find the volume of this cone.

Katen [24]

Answer:

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3 0

3 years ago

Maths!

Ber [7]

Answer:

(1) Variance = 4.5 and Standard deviation = 2.121.

(2) Variance = 4.5 and Standard deviation = 2.121.

(3) The effect on the measure of dispersion if each value is changed uniformly is that it remains unchanged.

Step-by-step explanation:

We are given with the following set of data below;

X $X-\bar X$ $(X-\bar X)^{2}$

5 5 - 8 = -3 9

8 8 - 8 = 0 0

10 10 - 8 = 2 4

9 9 - 8 = 1 1

6 6 - 8 = -2 4

Total 72 36

Firstly, the mean of the above data is given by;

Mean, $\bar X$ = $\frac{\sum X}{n}$

= $\frac{72}{9}$ = 8

(1)Now, the variance of the given data is;

Variance = $\frac{\sum (X-\bar X)^{2} }{n-1}$

= $\frac{36}{9-1}$ = 4.5

So, the standard deviation, (S.D.) = $\sqrt{\text{Variance}}$

= $\sqrt{4.5}$ = 2.12

(2) Now, each value is added by 2; so the new data set is given by;

X $X-\bar X$ $(X-\bar X)^{2}$

7 7 - 10 = -3 9

10 10 - 10 = 0 0

12 12 - 10 = 2 4

11 11 - 10 = 1 1

8 8 - 10 = -2 4

Total 90 36

Firstly, the mean of the above data is given by;

Mean, $\bar X$ = $\frac{\sum X}{n}$

= $\frac{90}{9}$ = 10

(1)Now, the variance of the given data is;

Variance = $\frac{\sum (X-\bar X)^{2} }{n-1}$

= $\frac{36}{9-1}$ = 4.5

So, the new standard deviation, (S.D.) = $\sqrt{\text{Variance}}$

= $\sqrt{4.5}$ = 2.12

(3) The effect on the measure of dispersion if each value is changed uniformly is that it remains unchanged as we see in the case of variance or standard deviation.

4 0

3 years ago

An integer is three more than another integer. Twice the larger integer is two less

Amiraneli [1.4K]

Answer:

The integers are 4 and 7 or -2 and 1.

Step-by-step explanation:

You can make a system of equations with the description of the two integers.

1. x = y + 3

2. 2x + 2 = y^2

The simplest and the fastest way to solve this system in this case is substitution. You can substitute x for y + 3 in the second equation.

1. x = y + 3

2. 2(y + 3) + 2 = y^2

Now simplify and solve the second one. For convenience, I will just disregard the first equation for now.

2y + 6 + 2 = y^2

y^2 - 2y - 8 = 0

You can factor this equation to solve for y.

(y - 4) (y + 2) = 0

y = 4, y = -2

Now we can substitute the value of y for x in the first equation.

x = 7, x = 1

3 0

3 years ago

(a) Find a vector parallel to the line of intersection of the planes −4x+2y−z=1 and 3x−2y+2z=1.

valentinak56 [21]

Find the intersection of the two planes. Do this by solving for z in terms of x and y ; then solve for y in terms of x ; then again for z but only in terms of x.

-4x + 2y - z = 1 ==> z = -4x + 2y - 1

3x - 2y + 2z = 1 ==> z = (1 - 3x + 2y)/2

==> -4x + 2y - 1 = (1 - 3x + 2y)/2

==> -8x + 4y - 2 = 1 - 3x + 2y

==> -5x + 2y = 3

==> y = (3 + 5x)/2

==> z = -4x + 2 (3 + 5x)/2 - 1 = x + 2

So if we take x = t, the line of intersection is parameterized by

r(t) = ⟨t, (3 + 5t )/2, 2 + t⟩

Just to not have to work with fractions, scale this by a factor of 2, so that

r(t) = ⟨2t, 3 + 5t, 4 + 2t⟩

(a) The tangent vector to r(t) is parallel to this line, so you can use

v = dr/dt = d/dt ⟨2t, 3 + 5t, 4 + 2t⟩ = ⟨2, 5, 2⟩

or any scalar multiple of this.

(b) (-1, -1, 1) indeed lies in both planes. Plug in x = -1, y = 1, and z = 1 to both plane equations to see this for yourself. We already found the parameterization for the intersection,

r(t) = ⟨2t, 3 + 5t, 4 + 2t⟩

3 0

3 years ago