Find the 60th term of the arithmetic sequence -27, -24, -21, ...

Mrs. Jacobs buys 20 kilograms if rice at $0.84 per kilogram. she buys 700 grams of shrimp at $1.02 per 100 grams.

rusak2 [61]

0.84(20)=$16.80
700/100=7
7(1.02)=$7.14
$7.14+16.80=$23.94 in total

8 0

3 years ago

Help please solve<br> <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1

Shkiper50 [21]

Answer:

$\displaystyle -\frac{1}{2} \leq x < 1$

Step-by-step explanation:

<u>Inequalities</u>

They relate one or more variables with comparison operators other than the equality.

We must find the set of values for x that make the expression stand

$\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0$

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

$\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0$

Simplifying by x-1 and taking x=1 out of the possible solutions:

$\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0$

We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions

$(6x^3+14x^2+10x+12)$

is always positive and doesn't affect the result. It can be neglected. The expression

$(x-\frac{1}{2})^2$

can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression

$\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0$

For the expression to be negative, both signs must be opposite, that is

$(x+\frac{1}{2})\geq 0, (x-1)$

Or

$(x+\frac{1}{2})\leq 0, (x-1)>0$

Note we have excluded x=1 from the solution.

The first inequality gives us the solution

$\displaystyle -\frac{1}{2} \leq x < 1$

The second inequality gives no solution because it's impossible to comply with both conditions.

Thus, the solution for the given inequality is

$\boxed{\displaystyle -\frac{1}{2} \leq x < 1 }$

7 0

3 years ago

Please? I need help here

Yuki888 [10]

Answer:

Step-by-step explanation:

If pair of ratios does form a true proportion then:

$a:b=c:d\to\dfrac{a}{b}=\dfrac{c}{d}\to ad=bc$

$A)\\\\8:14=20:35\to\dfrac{8}{14}=\dfrac{20}{35}\\\\(8)(35)=(14)(25)\\\\280=280\qquad\bold{TRUE}\\\\b)\\\\\dfrac{6}{10}=\dfrac{15}{25}\\\\(6)(25)=(10)(15)\\\\150=150\qquad\bold{TRUE}$

$c)\\\\\dfrac{9}{5}=\dfrac{27}{15}\\\\(9)(15)=(5)(27)\\\\135=135\qquad\bold{TRUE}\\\\d)\\\\12:15=30:50\to\dfrac{12}{15}=\dfrac{30}{50}\\\\(12)(50)=(15)(30)\\\\600=450\qquad\bold{FALSE}$