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zhannawk [14.2K]
2 years ago
13

Raja purchased vegetables for Rs. 98.50, grocery for Rs. 326.25 fruit for

Mathematics
1 answer:
ratelena [41]2 years ago
8 0

Answer:

https://brainly.in/question/6062336

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What is the mean (average) for -36, -25, 12, -4, 18?
Alina [70]
To find the mean add all the numbers up then divide by how many numbers you have.

-36+(-25)+12+(-4)+18= (-35)

there are a total of five numbers so divide by 5

(-35)/5=(-7)

answer -7
3 0
3 years ago
Read 2 more answers
In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= root 2 cm.
Nata [24]

Given CD is an altitude such that AD=BC , AB=3 cm and CD= √2 cm.

Let AD=x, Since given AB=3

                                     AD+DB=3

                                       x+DB = 3

                                        DB = 3-x

Since ΔBCD is rght angle triangle, let's apply Pythagoras theorem

BC^{2} = DB^{2} +CD^{2}

BC^{2} = (3-x)^{2} +(\sqrt{2} )^{2}

BC^{2} =(3-x)^{2} +2

Since given AD=BC,let us plugin BC=x in above step.

x^{2} =(3-x)^{2} +2

x^{2} =9-6x+x^{2} +2

6x=11

x=\frac{11}{6}

Now we know AD=x=\frac{11}{6} and given CD=√2.

Let us apply Pythagoras theorem for ΔACD

AC^{2} =AD^{2} +DC^{2}

AC^{2} = (\frac{11}{6} )^{2} +(\sqrt{2} )^{2}

AC^{2} =\frac{193}{36}

AC=\sqrt{\frac{193}{36} } = 2.315cm

3 0
3 years ago
Can you pls help me thank you have a good day
expeople1 [14]

Answer:

the slope is  3/1  so if the point before your answer is (2,6) the add 3 to six and 1 to 2 and your answer is (3,9).

hope this helps

8 0
2 years ago
Read 2 more answers
A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

7 0
2 years ago
Given m||n, find the value of x.<br> +<br> 360<br> -Po
AVprozaik [17]

Answer:

Step-by-step explanation:

13.6

6 0
2 years ago
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