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3 years ago

700 students attend Ridgewood Junior High School. 5% of students bring tlieir lunch

Mathematics

1 answer:

Butoxors [25]3 years ago

6 0

Answer:

Step-by-step explanation:

bc if u change the 5% into .05 than u multiply it to get 35 sorry if I am wrong or to late

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What is the radius of the garden?

shutvik [7]

Step-by-step explanation:

Area of circle= Πr²

Πr² = 132² m

r²=132²/Π m

=5546.23m

r =√5546.23

=74.473 m

5 0

2 years ago

Determine if equation has one solution ,no solution ,or an infinite number of solutions justify the answer

tangare [24]

Answer:

no solution

Step-by-step explanation:

- 4(x + 2) = - 4x - 5 ← distribute parenthesis on left side by - 4

- 4x - 8 = - 4x - 5 ( add 8 to both sides )

- 4x = - 4x + 3 ( add 4x to both sides )

0 = 3 ← False statement

This indicates the equation has no solution

8 0

2 years ago

On a map, the distance between two towns is 2.5 inches, and 1 inch represents 6 miles. What is the actual distance between the t

White raven [17]

X=6.2*20
X=124 MILES BETWEEN THESE 2 TOWNS.
Hope this helps you:)

7 0

3 years ago

$34,300 loan at 3.5% for 4 months.

cestrela7 [59]

keeping in mind that 4 months is not even a year, since there are 12 months in a year, 4 months is then 4/12 years.

let's assume is simple interest.

$\bf ~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$34300\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ t=years\to \frac{4}{12}\dotfill &\frac{1}{3} \end{cases} \\\\\\ A=34300\left[ 1+(0.035)\left( \frac{1}{3} \right) \right]\implies A= 34300(1.011\overline{6})\implies A=34700.1\overline{6}$

5 0

3 years ago

Dy/dx if y = Ln (2x3 + 3x).

NeTakaya

Answer:

$\frac{6x^2+3}{2x^3+3x}$

Step-by-step explanation:

You need to apply the chain rule here.

There are few other requirements:

You will need to know how to differentiate $\ln(u)$ .

You will need to know how to differentiate polynomials as well.

So here are some rules we will be applying:

Assume $u=u(x) \text{ and } v=v(x)$

$\frac{d}{dx}\ln(u)=\frac{1}{u} \cdot \frac{du}{dx}$

$\text{ power rule } \frac{d}{dx}x^n=nx^{n-1}$

$\text{ constant multiply rule } \frac{d}{dx}c\cdot u=c \cdot \frac{du}{dx}$

$\text{ sum/difference rule } \frac{d}{dx}(u \pm v)=\frac{du}{dx} \pm \frac{dv}{dx}$

Those appear to be really all we need.

Let's do it:

$\frac{d}{dx}\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot \frac{d}{dx}(2x^3+3x)$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (\frac{d}{dx}(2x^3)+\frac{d}{dx}(3x))$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (2 \cdot \frac{dx^3}{dx}+3 \cdot \frac{dx}{dx})$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (2 \cdot 3x^2+3(1))$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (6x^2+3)$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{6x^2+3}{2x^3+3x}$

I tried to be very clear of how I used the rules I mentioned but all you have to do for derivative of natural log is derivative of inside over the inside.

Your answer is $\frac{dy}{dx}=\frac{(2x^3+3x)'}{2x^3+3x}=\frac{6x^2+3}{2x^3+3x}$ .

3 0

3 years ago