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3 years ago

5

What’s is the range of the function

2 answers:

uranmaximum [27]3 years ago

7 0

Answer:

$y\leq 1$

Step-by-step explanation:

The range of a function is all the values that $y$ could take, in this graph the highest value for $y$ is 1, because of this the answer must include all values bellow 1.

Also, the value is marked with a filled point, this means $y$ takes this value, so the answer needs to contain $y=1$ , putting all together $y\leq 1$ .

jolli1 [7]3 years ago

3 0

Answer:

Y <= 1

Step-by-step explanation:

This is because it is a solid line and a filled point on the top line, the bottom line is irrelevant for this question

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Write an equation in point-slope form for the line that has a slope of 5 and contains the point (-1,2)

Cloud [144]

Answer:

y - 2 = 5(x+1)

Step-by-step explanation:

Point slope form is

y-y1 = m(x-x1) where m is the slope and ( x1,y1) is a point on the line

y - 2 = 5(x- -1)

y - 2 = 5(x+1)

4 0

3 years ago

Read 2 more answers

A golfer keeps track of his score for playing nine holes of golf (half a normal golf round). His mean score is 8080 with a sta

solniwko [45]

Answer:

Therefore the mean and standard deviation of his total score if he plays a full 18 holes are 160 and $11\sqrt2$ respectively.

Step-by-step explanation:

Given that,

For the first 9 holes X:

E(X) = 80

SD(X)=13

For the second 9 holes Y:

E(Y) = 80

SD(Y)=13

For the sum W=X+Y, the following properties holds for means , variance and standard deviation :

E(W)=E(X)+E(Y)

and

V(W)=V(X)+V(Y)

⇒SD²(W)=SD²(X)+SD²(Y) [ Variance = (standard deviation)²]

$\Rightarrow SD(W)=\sqrt{SD^2(X)+SD^2(Y)}$

∴E(W)=E(X)+E(Y) = 80 +80=160

and

∴ $SD(W)=\sqrt{SD^2(X)+SD^2(Y)}$

$=\sqrt{11^2+11^2}$

$=\sqrt{2.11^2}$

$=11\sqrt2$

Therefore the mean and standard deviation of his total score if he plays a full 18 holes are 160 and $11\sqrt2$ respectively.

4 0

3 years ago

Plllzzz im new and i neeed help

ella [17]

Answer:

4082

Step-by-step explanation:

Given

The composite object

Required

The volume

The object is a mix of a cone and a hemisphere

Such that:

<u>Cone</u>

$r = 10cm$ ---- radius (r = 20/2)

$h = 19cm$

<u>Hemisphere</u>

$r=10cm$

The volume of the cone is:

$V_1 = \frac{1}{3}\pi r^2h$

$V_1 = \frac{1}{3}\pi * 10^2 * 19$

$V_1 = \frac{1900}{3}\pi$

The volume of the hemisphere is:

$V_2 = \frac{2}{3}\pi r^3$

$V_2 = \frac{2}{3}\pi 10^3$

$V_2 = \frac{2000}{3}\pi$

So, the volume of the object is:

$V = V_1 + V_2$

$V = \frac{1900}{3}\pi + \frac{2000}{3}\pi$

$V = \frac{3900}{3}\pi$

$V = 1300\pi$

$V = 1300 * 3.14$

$V = 4082$

8 0

3 years ago

each month for 7 months, samuel mwos 3 lawns. How many more lawns does he need to mow before he has mowed 29 laws?

ira [324]

He needs to mow 8 more lawns

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3 years ago

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If sectheta=13/12 and 270°<theta<360°, then tantheta=

Tpy6a [65]

Answer:

this should be 450° hahah sorry if I am wrong I don't know!

5 0

3 years ago