All categories

2 years ago

Plz help im tired ty

Mathematics

1 answer:

saul85 [17]2 years ago

6 0

Answer:

Hi, your answer would be -4/27

Step-by-step explanation:

You just divided negative 1/9 by 3/4 to get -4/27.

You might be interested in

Find the terms, coefficients, and constants of 4x+x+17

andreyandreev [35.5K]

Answer:

17 and x=constant,

4x=coefficient

hope this helps

have a good day :)

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

15.24 x 1.64 the answer needs to be converted from us dollars to British pounds and can i have steps to plz

olga2289 [7]

Answer:

19.334

Step-by-step explanation:

First I multiplied.

15.24 x 1.64 = 24.9936

Then I convert my answer to British pounds.

24.9936 = 19.334 British Pound Sterling

Have a blessed day and hope this helps.

7 0

3 years ago

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

Is it possible to create two triangles with congruent side lengths but different angle measurements?

kotykmax [81]

Yes. If the side lengths are different, you can end up with different angle measurements (example: SSA~ property. You can have two sides that are the same but you can make two different triangles with those side lengths and that one angle.)

8 0

2 years ago

I SPENT 50 POINTS ON THIS, PLEASE HELP

Anestetic [448]

In the figure, we can consider that the base is the side that mesures 12 in and that the height is the side that measures 15 in, since that sides are perpendicular. So, we just need to use the given formula:

$A = \dfrac{bh}{2}\Longrightarrow A = \dfrac{12\cdot15}{2}\Longrightarrow A=6\cdot15\Longrightarrow \boxed{A = 90~\text{in}^2}$

Hence, the area of the triangle is 90 in² (B).

5 0

3 years ago

Read 2 more answers