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3 years ago

PLS HELP!!!

Mathematics

1 answer:

satela [25.4K]3 years ago

7 0

A=180-105
A=75

In the triangle
75+ 9x +12x =180
X=5

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5417 times 2.4 equals what

Andre45 [30]

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3 years ago

Sophie has to choose seven different positive (non zero) whole numbers whose mean is 7.

Anna71 [15]

Answer:

The largest possible number is 28.

Step-by-step explanation:

Sophie has to choose seven different numbers whose mean is 7, this means that the sum of the numbers is 49. (since the mean is the sum of them all divided by 7 and if we want the mean to be 7, then the sum would need to be 49).

We are asked what is the largest possible number she could choose, so therefore, the other six numbers should be non-zero and the smallest ones so she would have to choose the numbers from 1 to 6 first.

Let's see how much they sum up: $1+2+3+4+5+6=21$

So now, she needs one more number to sum up to 21 and that will give her 49 in total, this number is 28 (since 28 + 21 =49).

Thus, 28 is the largest number she could choose because if she chose any other greater number the mean would be bigger than 49.

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3 years ago

Janaya buys a postcard from each travel destination she visits. Before starting her travels this year, Janaya had 20 postcards.

MrRa [10]

Answer:

The percentage increase in Jayana's postcard collection is 15%

Step-by-step explanation:

Jayana had 20 postcards last year

With 3 more postcards this year

From the given equation Quantity = Percent x Whole

Here;

Let P = Percent/100%

Quantity = Increment in postcard = 3 postcards

Whole = The total postcards from last year = 20 postcards

Hence,

3 postcards = P x 20 postcards

P = 3/20 = 0.15

Recall P = Percent/100%

∴ Percent = P x 100% = 0.15 x 100% = 15%

The percentage increase in Jayana's postcard collection is 15%

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3 years ago

If 1 kg of milk had 0.264 kg of fat, then find the amount fat in 12.5 kg of milk.<br>

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1 kg--> 0.264 kg fat

:. 12.5 kg--> (0.264*12.5) 3.3 kg fat

Mark me brainliesttt :))

3 0

3 years ago

Find the absolute minimum and absolute maximum values of f on the given interval. f(x) = (x2 − 1)3, [−1, 2]

Alja [10]

Given :

$F(x)=3(x^2-1)$

To Find :

the absolute minimum and absolute maximum values of f on the given interval

[-1,2] .

Solution :

Now , getting first order differential equation and equating its equal to zero.

$\dfrac{d(3x^2-3)}{dx}=0\\\\6x=0\\x=0$

So , x=0 is critical point .

Now , coefficient of $x^2$ is positive .

Therefore , it is increasing function after x=0 .

So , min value will be at , x=0.

$Min = (0^2-1)\times 3=-3$

And maximum value will be the maximum at the x=2 because it is increasing function .

$Max=(2^2-1)\times 3=9$

Therefore , max and min value is 9 and -3 respectively .

Hence , this is the required solution .

3 0

3 years ago