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antoniya [11.8K]
3 years ago
7

Does anyone know the answer ?

Mathematics
2 answers:
Alecsey [184]3 years ago
5 0

Answer:

As I’m unsure as to whether or not this is a joke or not, I’ll say 15 cm.

ycow [4]3 years ago
3 0
It’s 15cm for the answer
You might be interested in
What is the decimal expansion of 11/40
Katena32 [7]

Answer:

0.275

Step-by-step explanation:

5 0
3 years ago
A student’s score on an exam is an integer between 0 and 100. A score of 90 to 100 is an A, 80 to 89 is a B, 70 to 79 is a C, 60
elixir [45]

Answer:

A. 0.22

B. 0.18

C. 0.25

D. 0.244

Step-by-step explanation:

S = {51 to 100} = 50

The sample space S contains values from 51 to 100 which is a total of 50 different values.

A.

Probability of A (lies between the values of 90 to 100 = 11).

11/50 = 0.22

B.

For a student to fail the course, his course has to be less than 60 = from 51 to 59. A total of 9 values.

9/50 = 0.18

C.

For student to get c, (70 to 79) a total of 10 values: 10/50 = 0.20

P(student did not get C) = 1-0.20 = 0.80

To get B, ( 80 to 89)

10/50 = 0.20

Probability that a student who is known not to have a c grade has a b grade = 0.20/0.80 = 0.25

D.

Probability of passing lies between 60 to 100 = 41 scores

41/50 = 0.82

Probability of student who passed having a B = 0.20/0.82 = 0.244

4 0
2 years ago
A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient
ankoles [38]

<u>Answer:</u>

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

<u>Solution:</u>

We need to show that the gradient of the curve at A is 1

Here given that ,

y=(x-a) \sqrt{(x-b)}  --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

0=(b+1-a) \sqrt{(b+1-b)}

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

y^{\prime}=u^{\prime} y+y^{\prime} u

so, we get

{u}^{\prime}=1

v^{\prime}=\frac{1}{2} \sqrt{(x-b)}

y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}

By putting value of point A and putting value of eq 2 we get

y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}

y^{\prime}=\frac{d y}{d x}=1

Hence proved that the gradient of the curve at A is 1.

7 0
2 years ago
Is anyone able to answer this for me? GIVING BRAINLIEST! hurry!
svetoff [14.1K]
Nicoles pattern:

1
5
17
53
161

Ian’s pattern:

0
1
3
7
15

Ordered pair:

(1, 0)
(5, 1)
(17, 3)
(53, 7)
(161, 15)



Table 1 -

Sequence 1:

9
11
13
15
17

Sequence 2:

5
8
11
14
17

Ordered pair:

(9, 5)
(11, 8)
(13, 11)
(15, 14)
(17, 17)

Table 2 -

Sequence 1:

20
16
12
8
4

Sequence 2:

20
17
14
11
8

Ordered pair:

(20, 20)
(16, 17)
(12, 14)
(8, 11)
(4, 8)

Table 3 -

Sequence 1:

1
3
7
15
31

Sequence 2:

40
24
16
12
10

Ordered pair:

(1, 40)
(3, 24)
(7, 16)
(15, 12)
(31, 10)
5 0
2 years ago
20 POINTS!!!!!!
Crazy boy [7]
You will need three roots for this, so we have
Let x = -30, -10 and +20
So the factors will be (x+30)(x+10)(x-20)
The divide it to 100, this will help bring the peak up and down
So the polynomial function R(x) will become
1/100 * (x+30)(x+10)(x-20)
R(x) = 1/100 * (x+30)(x+10)(x-20)

Finding the X-intercept:
Let R(x) = 0 and solve for x.
1/100 * (x+30)(x+10)(x-20) = 0
x = -30, -10, 20 are the x-intercepts.

8 0
3 years ago
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