I'll give you brainiest and lots of points..

Set up a system of equations needed to solve each probler

goblinko [34]

Answer:

This is not a solvable question, since the least amount of money you can have with this combination is $4.20.

Step-by-step explanation:

Let q = quarters

Let d = dimes

Equation: 0.1d + 0.25q = 3.00

Maximum amount of dimes: 0.1d *42 = 4.2d

8 0

4 years ago

Write an algebraic expression for the word expression. the quotient of 10 times m and 7

Makovka662 [10]

Answer:

10m/7

Step-by-step explanation:

quotient means division so 10 times m is 10m

and then we will divide 10m and 7

10m/7

please mark as brainliest

8 0

3 years ago

Peter is at a lumber yard. He gets 2 free boxes of nails for every 10 boards he buys. Write an expression for the number of boxe

Sati [7]

A. The number of 10-boards Peter bought is equal to n divided by 10. Then, each of the 10-boxes will get two boxes of nails. The number of boxes of nails that Peter will have after buying n boards will be,

N = (2)(n/10)

Simplifying,

<em> N = n/5</em>

b. If the number of boards are 90 then,

N2 = (90/10)(2)(100 nails/box)
N2 = 1800

Answer: 1800

7 0

4 years ago

Read 2 more answers

PLEASE HELP! 13 POINTS! A fitness center has two membership plans. One has a low dollar sign up fee of 15 dollars and costs 38 d

il63 [147K]

Answer:

X = 9 months

Step-by-step explanation:

To solve the number of months when the cost of each payment

are the same, you should equate the two cost. First establish the equations

Let A be the cost of first payment method

B is the cost of the second payment method

X is the number of months

A = 38x + 15

B = 31x + 78

38x + 15 = 31x + 78

7x = 63

X = 9 months

5 0

4 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cint%20t%5E2%2B1%20%5C%20dt" id="TexFormula1" title="\frac{d}{dx} \

Kisachek [45]

Answer:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2$

Step-by-step explanation:

$\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} = \ ?$

We can use Part I of the Fundamental Theorem of Calculus:

$\displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}$

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

$\displaystyle\int\limits^b_a \text{f(t) dt} + \int\limits^c_b \text{f(t) dt} = \int\limits^c_a \text{f(t) dt}$

We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

$\displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

$\displaystyle\int\limits^b_a \text{f(t) dt}\ = -\int\limits^a_b \text{f(t) dt}$

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

$\displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}$

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

$\displaystyle \frac{d}{dx} \int\limits^u_a \text{f(t) dt} = \text{f(u)} \cdot \frac{d}{dx} [u]$
where u represents any function other than a variable

For the first term, replace $\text{t}$ with $2x$ , and apply the chain rule to the function. Do the same for the second term; replace

$\displaystyle-[(2x)^2+1] \cdot (2) \ + \ [(x^2)^2 + 1] \cdot (2x)$

Simplify the expression by distributing $2$ and $2x$ inside their respective parentheses.

$[-(8x^2 +2)] + (2x^5 + 2x)$
$-8x^2 -2 + 2x^5 + 2x$

Rearrange the terms to be in order from the highest degree to the lowest degree.

$\displaystyle2x^5-8x^2+2x-2$

This is the derivative of the given integral, and thus the solution to the problem.

6 0

3 years ago