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2 years ago

Can someone help me with this plz

Mathematics

2 answers:

k0ka [10]2 years ago

6 0

56 would be the answer

allochka39001 [22]2 years ago

5 0

Okay so any calculator works:

2 + 2 + 6 + 6 = 16

16 * 3.5 = 56 cm.

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5) ASAP HELP DUE IN 5MIN

Sever21 [200]

The answer is y=-5x :)

7 0

2 years ago

Read 2 more answers

The dimensions of the tank are 50 cm by 32 cm by 20 cm. The tank is full of water and sand. The ratio of the volume of water to

Sauron [17]

Given:

Dimensions of a tank are 50 cm by 32 cm by 20 cm

The tank is full of water and sand

Ratio of volume of water to the volume of sand is 5:1

To find:

Volume of water in the tank in liters

Solution:

Without loss of generality, let,

Length of tank = 50 cm

Breadth of tank = 32 cm

Height of tank = 20 cm

Then, we have,

Total Volume of the tank = Length * Breadth * Height

$\Rightarrow$ Total Volume = 50*32*20 cc

$\Rightarrow$ Total Volume = 32000 cc

It is given that,

Volume of water : Volume of sand = 5:1

$\Rightarrow$ (Volume of water)/(Volume of sand) = 5/1

Cross multiplying, we have,

Volume of sand = (Volume of water)/5

It is also given that the tank is full of water and sand. This implies that,

Volume of water + Volume of sand = Total Volume

Using the value, we obtained, we get,

Volume of water + (Volume of water)/5 = Total Volume

$\Rightarrow$ (6 * Volume of water)/5 = Total Volume

Plugging in the value of total volume, we get,

$\Rightarrow$ (6 * Volume of water)/5 = 32000 cc

$\Rightarrow$ Volume of water = $\frac{32000*5}{6}$ cc

$\Rightarrow$ Volume of water $\approx$ 26,666.6667 cc

We know that,

1 cc = 0.001 liters

$\Rightarrow$ 26,666.6667 cc = (26666.6667 * 0.001) liters

$\Rightarrow$ 26,666.6667 cc $\approx$ 26.667 liters

Thus, Volume of water in the tank is approximately 26.667 liters

Final Answer:

Volume of water in the tank (in liters) is approximately 26.667 liters

4 0

2 years ago

How much simple interest rate would earn if had $10,000 and you were being paid 5% for 15 months?

lakkis [162]

Here is the interest equation: I=prt
Where I is interest, p is principal, r is rate and t is time(in years).
Substitute with known values.

I= prt
I= 10000(.05)(1 and 1 over 4)
I= 625
Answer: $625 for 15 months.

4 0

3 years ago

What is 8.7 divided by 53.07 step by step?

tresset_1 [31]

Shift the decimal points of both 8.7 and 53.07 to the right by 2 decimal places.

(This is needed to convert 53.07 to 5307, since long division must divide by a whole number.)

870/5307

Set up the long division.

5307|870

Therefore your answer should be 0.1639.

Hope I help! Please check my questions please. I need major help on the question.

6 0

3 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago