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3 years ago

14

Pls I need help please :D its math and you'll get 30 points Its a picture

1 answer:

Amanda [17]3 years ago

5 0

Answer:

5,750

Step-by-step explanation:

if he has 10 of each then you can multiply each bag by 10 and then multiply the amount of bags to the amount the coin is worth

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Solve the proportion = 2x+?<br> 2x+2 for x.

bonufazy [111]

No proportion here. Repost your question.

3 0

3 years ago

Does anyone know how to do this I need help please

Mkey [24]

Answer:

Step-by-step explanation:

4 0

3 years ago

4x^2+2x-5-3x and 6x^2-x+3-2x^2-8<br><br> Are these two expressions equal? Show how you know

aivan3 [116]

Step-by-step explanation:

Simply Equation 1

R. H. S

6x^2 -x +3-2x^2-8

==> 6x^2-2x^2 -x +3-8

==> 4x^2-x-5 .....(1)

Simply equation 2

L. H. S

4x^2+2x-5-3x

==> 4x^2 +2x-3x-5

==> 4x^2-x-5 ....(2)

Since here, RHS = LHS

both eqn (1) = eqn(2)

hence they are equal.

7 0

3 years ago

Read 2 more answers

There are 6 rotten mangoes and 24 good mangoes in a bag. What fraction of the mangoes is rotten?

dem82 [27]

$\sf \bf {\boxed {\mathbb {Given:}}}$

Total number of rotten mangoes in the bag = 6

Total number of good mangoes in the bag = 24

$\sf \bf {\boxed {\mathbb {To\:find:}}}$

Fraction of the rotten mangoes to the total mangoes.

$\sf \bf {\boxed {\mathbb {Solution:}}}$

$\implies {\blue {\boxed {\boxed {\purple {\sf { \frac{1}{5} }}}}}}$

$\sf \bf {\boxed {\mathbb {Step-by-step\:explanation :}}}$

Total number of mangoes in the bag = Total number of rotten mangoes + Total number of good mangoes

➺ $\:6 + 24$

➺ $\:30$

Now,

Fraction = $\frac{Total \: \: number \: \: of \: \: rotten \: \: mangoes}{Total \: \: number \: \: of \: \: mangoes}$

➺ $\: \frac{6}{30}$

➺ $\: \frac{1}{5}$

Therefore, the fraction of the rotten mangoes to the total mangoes is $\sf\pink{\frac{1}{5}}$ .

$\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{❦}}}}}$

6 0

3 years ago

Read 2 more answers

Use logarithmic differentiation to find the derivative of the function. y = sqrt(x)e^x^2(x^2 + 5)^12

padilas [110]

$y=\sqrt x\,e^{x^2}(x^2+5)^{12}=x^{1/2}e^{x^2}(x^2+5)^{1/2}$

Take the logarithm of both sides and expand the right hand side:

$\ln y=\ln\left(x^{1/2}e^{x^2}(x^2+5)^{1/2}\right)$

$\ln y=\ln x^{1/2}+\ln e^{x^2}+\ln(x^2+5)^{12}$

$\ln y=\dfrac12\ln x+x^2\ln e+12\ln(x^2+5)$

$\ln y=\dfrac12\ln x+x^2+12\ln(x^2+5)$

Now take the derivative of both sides with respect to $x$ :

$\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1{2x}+2x+\dfrac{24x}{x^2+5}$

$\dfrac{\mathrm dy}{\mathrm dx}=\left(\dfrac1{2x}+2x+\dfrac{24x}{x^2+5}\right)y$

$\dfrac{\mathrm dy}{\mathrm dx}=\left(\dfrac1{2x}+2x+\dfrac{24x}{x^2+5}\right)\sqrt x\,e^{x^2}(x^2+5)^{12}$

I'd stop there, but you could condense the right side a bit to get

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{4x^4+21x^2+48x+5}{2x(x^2+5)}\sqrt x\,e^{x^2}(x^2+5)^{12}$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{4x^4+21x^2+48x+5}{2\sqrt x}e^{x^2}(x^2+5)^{11}$

8 0

4 years ago