P-(2q+7) when p=10 and q=3

steposvetlana [31]

Draw an arrow pointing to the left on left side of the open circle

4 0

3 years ago

Kate prepared 9 kilograms of dough after working 3 hours. How many hours did Kate work if she prepared 12 kilograms of dough? As

beks73 [17]

Answer:

4 hours

Step-by-step explanation:

cross multiply

3 0

3 years ago

I have to right a statistic question would this be one Out of 60 7th graders 45 or 75% of them prefer to play sports after schoo

mina [271]

Answer:

This really isn't a question. I can help you out with writing one.

John randomly selected 45% of seventh grade students in his school and asked them what their favorite time to play sports was. Of the students surveyed, 51 prefer to play sports after school. Based on this data, what is the most reasonable prediction of seventh graders in his entire grade?

Step-by-step explanation:

To answer my question, you would set up a proportion.

$\frac{9}{20} = \frac{51}{x}$ <-- the answer to that question will will be 113.

5 0

3 years ago

From 1st principle find the derivative of log (ax+b)

OLga [1]

Answer:

Step-by-step explanation:

The parent function here is y = log x, where 10 is the base.

The derivative of y = log x is dy/dx = (ln x) / ln 10.

The derivative of y = log (ax+b) is found in that manner, but additional steps are necessary: differentiate the argument ax + b:

The derivative with respect to 10 of log (ax + b) is:

dy/dx = [ 1 / (ax + b) ] / [ ln 10 ] *a, where a is the derivative of (ax + b).

Alternatively, we could express the answer as

dy/dx = [ a / (ax + b) ] / [ ln 10 ]

5 0

3 years ago

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$

As I mentioned before, this volume is equal to 1, hence:

$\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }$

3 0

2 years ago

P-(2q+7) when p=10 and q=3​

P-(2q+7) when p=10 and q=3