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In-s [12.5K]
3 years ago
15

40 is how many times less than 400

Mathematics
2 answers:
lisov135 [29]3 years ago
6 0
400/40 = 10
..............
Akimi4 [234]3 years ago
6 0
40 is 10 times less than 400. to find this you divide 400 by 40
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What is the domain of (cd)(x)?
viktelen [127]

Answer:

Only option B is correct, i.e. all real values of x except x = 2.

Step-by-step explanation:

Given the functions are C(x) = 5/(x-2) and D(x) = (x+3)

Finding (C·D)(x) :-

(C·D)(x) = C(x) * D(x)

(C·D)(x) = 5/(x-2) * (x+3)

(C·D)(x) = 5(x+3) / (x-2)

(C·D)(x) = (5x+15) / (x-2)

Let y(x) = (C·D)(x) = (5x+15) / (x-2)

According to definition of functions, the rational functions are defined for all Real values except the one at which denominator is zero.

It means domain will be all Real values except (x-2)≠0 or x≠2.

Hence, only option B is correct, i.e. all real values of x except x = 2.

6 0
3 years ago
Given tanx=8/15 when pi < x < 3pi/2, find tan x/2.
S_A_V [24]

Answer:

tan(x/2)=-4

Step-by-step explanation:

we know that

The angle x belong to the III quadrant -----> given problem

step 1

Find the angle x

x=arctan(8/15)=28.07\°

Remember that the angle x belong to the III Quadrant

so

x=28.07\°+180\°=208.07\°

step 2

Find angle x/2

x/2=208.07\°/2=104.035\°

step 3

Find tan(x/2)

tan(104.035\°)=-4

8 0
3 years ago
How can you write the equation for a linear function if you know only two ordered pairs for the function?
dedylja [7]

Answer:

See below

Step-by-step explanation:

We first have to understand the parts of a y=mx+b equation

y=mx+b

m: Is the slope

b: Is the y-intercept

x: is the x part of (x,y)

y: is the y part of(x,y)

So...

if we use an example ordered pair like (1,3)

and a slope of 2

We can plug everything into y=mx+b to get

1=2*3+b which simplifies to  b= -5

Now, you can use this concept to fill in the blanks below:  

Use the two ordered pairs to find the slope, (m).

Then substitute the slope, and (one set of ordered pairs) into y=mx+b to solve for (b).

6 0
3 years ago
I have trouble dividing​
emmasim [6.3K]

Answer:

I'm Sorry, do you have a question you need help with?

Step-by-step explanation:

8 0
3 years ago
One urn contains one blue ball (labeled B1) and three red balls (labeled R1, R2, and R3). A second urn contains two red balls (R
marusya05 [52]

Answer:

(a) See attachment for tree diagram

(b) 24 possible outcomes

Step-by-step explanation:

Given

Urn\ 1 = \{B_1, R_1, R_2, R_3\}

Urn\ 2 = \{R_4, R_5, B_2, B_3\}

Solving (a): A possibility tree

If urn 1 is selected, the following selection exists:

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

If urn 2 is selected, the following selection exists:

B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]

<em>See attachment for possibility tree</em>

Solving (b): The total number of outcome

<u>For urn 1</u>

There are 4 balls in urn 1

n = \{B_1,R_1,R_2,R_3\}

Each of the balls has 3 subsets. i.e.

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

So, the selection is:

Urn\ 1 = 4 * 3

Urn\ 1 = 12

<u>For urn 2</u>

There are 4 balls in urn 2

n = \{B_2,B_3,R_4,R_5\}

Each of the balls has 3 subsets. i.e.

B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]

So, the selection is:

Urn\ 2 = 4 * 3

Urn\ 2 = 12

Total number of outcomes is:

Total = Urn\ 1 + Urn\ 2

Total = 12 + 12

Total = 24

5 0
3 years ago
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