All categories

4 years ago

Ten increased by the quotient of a number and three is equal to nineteen. Find the number.

1 answer:

3 0

$\bf \stackrel{\textit{10 increased by }}{10 + }\stackrel{\textit{a quotient of \underline{x} and 3}}{\cfrac{x}{3}}~~\stackrel{equals}{=}~~19 \implies \cfrac{x}{3}=9\implies \boxed{x=27}$

You might be interested in

A piece of cardboard measures 10 ft by 10 ft. Four equal squares of size x are removed from the corners. After removing the squa

Gnesinka [82]

Answer:

The value of x would be $\frac{5}{3}$

Step-by-step explanation:

Given,

The dimension of the cardboard = 10 ft by 10 ft,

∵ After removing four equal squares of size x ( in ft ) from the corners,

The dimension of the resultant box would be,

Length = ( 10 - 2x ) ft,

Width = ( 10 - 2x ) ft,

Height = x ft,

The volume of box,

$V=(10-2x)\times (10 - 2x)\times x=x(10-2x)^2 = x(100 - 40x + 4x^2)=100x - 40x^2 + 4x^3$

Differentiating with respect to x,

$V'=100 - 80x + 12x^2$

Again differentiating with respect to x,

$V''=-80 + 24x$

For maxima or minima,

$V'=0$

$\implies 100 - 80x + 12x^2 = 0$

$\implies 3x^2 - 20x + 25=0$

By quadratic formula,

$x=\frac{20\pm \sqrt{20^2-4\times 3\times 25}}{6}$

$x=\frac{20\pm \sqrt{400 - 300}}{6}$

$x=\frac{20\pm \sqrt{100}}{6}$

$x=\frac{20\pm 10}{6}$

$\implies x = \frac{10}{6}=\frac{5}{3}\text{ or } x = 5$

For x = 5/3, V'' = negative,

While for x = 5, V'' = Positive,

Hence, the value of x would be 5/3 ft for maximising the volume.

8 0

4 years ago

Could someone please help me with this? Solution set of lx^2+mx+n=0 is...?

IgorLugansk [536]

Answer:

Simplifying

lx2 + mx + n = 0

Solving

lx2 + mx + n = 0

Solving for variable 'l'.

Move all terms containing l to the left, all other terms to the right.

Add '-1mx' to each side of the equation.

lx2 + mx + -1mx + n = 0 + -1mx

Combine like terms: mx + -1mx = 0

lx2 + 0 + n = 0 + -1mx

lx2 + n = 0 + -1mx

Remove the zero:

lx2 + n = -1mx

Add '-1n' to each side of the equation.

lx2 + n + -1n = -1mx + -1n

Combine like terms: n + -1n = 0

lx2 + 0 = -1mx + -1n

lx2 = -1mx + -1n

Divide each side by 'x2'.

l = -1mx-1 + -1nx-2

Simplifying

l = -1mx-1 + -1nx-2

Step-by-step explanation:

Hope this helped you!

6 0

4 years ago

Thank you for your help.

Nonamiya [84]

Answer:

q = -15

Step-by-step explanation:

5q = -75

Divide both sides by 5 : q = -15

8 0

2 years ago

Pleasantburg has a population growth model of P(t)=at2+bt+P0 where P0 is the initial population. Suppose that the future populat

yulyashka [42]

Answer:

The population will reach 34,200 in February of 2146.

Step-by-step explanation:

Population in t years after 2012 is given by:

$P(t) = 0.8t^{2} + 6t + 19000$

In what month and year will the population reach 34,200?

We have to find t for which P(t) = 34200. So

$P(t) = 0.8t^{2} + 6t + 19000$

$0.8t^{2} + 6t + 19000 = 34200$

$0.8t^{2} + 6t - 15200 = 0$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$0.8t^{2} + 6t - 15200 = 0$

So $a = 0.8, b = 6, c = -15200$

Then

$\bigtriangleup = 6^{2} - 4*0.8*(-15100) = 48356$

$t_{1} = \frac{-6 + \sqrt{48356}}{2*0.8} = 134.14$

$t_{2} = \frac{-6 - \sqrt{48356}}{2*0.8} = -141.64$

We only take the positive value.

134 years after 2012.

.14 of an year is 0.14*365 = 51.1. The 51st day of a year happens in February.

So the population will reach 34,200 in February of 2146.

6 0

3 years ago

PLEASE help! Functions problem!

bezimeni [28]

Answer:

Step-by-step explanation:

x = -1 and -7

6 0

3 years ago