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3 years ago

Please help ASAP!

Mathematics

2 answers:

marusya05 [52]3 years ago

8 0

Answer:

Not a function

Step-by-step explanation:

This is not a function because there are different outputs for the same input.

ad-work [718]3 years ago

8 0

Answer:

NOT A FUNCTION

Step-by-step explanation:

a function cant have multiple y coordinates for the same X

∞¬

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A cylindrical well is 20 meters deep and has a diameter of 1.5 meters. Approximately how many cubic meters of soil were dug out

madam [21]

V = πr2h

V = 3.14 x 0.75 x 0.75 x 20

V = 35.33 cubic meters

Hope i will become the brainilest.

6 0

3 years ago

What is the length of AB?

makkiz [27]

The radius of the circle is given to be r = 14.5

Let the center be O. So OB is the radius = 14.5

We can draw a triangle as shown in the image below. We have a right angled triangle. We know the hypotenuse and the base, we need to find the perpendicular side. This can be done using the Pythagorean theorem.

So, we can write:

$(14.5)^{2}=10^{2}+OA^{2} \\ \\ 
OA^{2}=110.25 \\ \\ 
OA=10.5$

Thus, the measure of OA = 10.5

OB = OA + AB

14.5 = 10.5 + AB
⇒
AB = 4

3 0

2 years ago

Read 2 more answers

Please could you find the answers to the questions in the attachment.

Fudgin [204]

( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ )we need 3 equations
1. midpoint equation which is ( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ ) when you have 2 points

2. distance formula which is D= $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height

so

x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=( $\frac{11+19}{2} , \frac{10+6}{2}$ )
midpoint=( $\frac{30}{2} , \frac{16}{2}$ )
midpoint= (15,8)

point x=(15,8)

y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=( $\frac{5+21}{2} , \frac{8+0}{2}$ )
midpoint=( $\frac{26}{2} , \frac{8}{2}$ )
midpoint=(13,4)

Y=(13,4)

legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= $\sqrt{(19-11)^{2}+(6-10)^{2}}$
D= $\sqrt{(8)^{2}+(-4)^{2}}$
D= $\sqrt{64+16}$
D= $\sqrt{80}$
D= $4 \sqrt{5}$
BC= $4 \sqrt{5}$

X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= $\sqrt{(13-15)^{2}+(4-8)^{2}}$
D= $\sqrt{(-2)^{2}+(-4)^{2}}$
D= $\sqrt{4+16}$
D= $\sqrt{20}$
D= $2 \sqrt{5}$
XY= $2 \sqrt{5}$

the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC= $4 \sqrt{5}$ and XY= $2 \sqrt{5}$

AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= $\sqrt{(21-5)^{2}+(0-8)^{2}}$
D= $\sqrt{(16)^{2}+(-8)^{2}}$
D= $\sqrt{256+64}$
D= $\sqrt{320}$
D= $4 \sqrt{2}$
AD= $4 \sqrt{2}$

so we have
AD= $4 \sqrt{2}$
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$

AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
( $4 \sqrt{2}+4 \sqrt{5}$ ) times 1/2 times $2 \sqrt{5}$ =
( $4 \sqrt{2}+4 \sqrt{5}$ ) times \sqrt{5} [/tex] =
$4 \sqrt{10}+4*5$ = $4 \sqrt{10}+20$ =80 $\sqrt{10}$ =252.982

X=(15,8)
Y=(13,4)
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$
Area=80 $\sqrt{10}$ square unit or 252.982 square units

7 0

3 years ago

Which of the following points is in the solution set of y < x2 - 2x - 8? (-2, -1)(0, -2)(4, 0)

ratelena [41]

Y < (x - 4)(x + 2)

so the critical points are -2 and 4
(-2,-1) will be in the solution

( 0,-2) will not

(4,0) will not.
the answer is ((-2,-1)

5 0

3 years ago

Read 2 more answers

What is the range for 25,25,21,28,37,26

lina2011 [118]

Answer:

16 is the range

Step-by-step explanation:

First, arrange the numbers from least to greatest.

21, 25, 25, 26, 28, 37

Then, subtract 21 (smallest number) from the 37 (biggest number).

37 - 21 = 16

So, the range is 16

Have a wonderful day!

7 0

2 years ago