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3 years ago

A solution to the equation x^3 + 2x - 25 = 0 lies between 2 and 3. Find this solution correct to 1 decimal place

Mathematics

1 answer:

Viktor [21]3 years ago

3 0

I have no idea??????

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Use matrix addition to solve this equation

REY [17]

Answer:

c11 = 8
c31 = 20
c22 = 14

Step-by-step explanation:

The sum of two matrices is the sum of corresponding terms.

$\left[\begin{array}{ccc}3&1&0\\-1&2&4\\9&7&-2\end{array}\right] +\left[\begin{array}{ccc}5&2&4\\1&12&3\\11&3&-2\end{array}\right] =\left[\begin{array}{ccc}8&3&4\\0&14&7\\20&10&-4\end{array}\right]$

8 0

3 years ago

Read 2 more answers

What does x= in the equation?

vampirchik [111]

Answer:

3.5

Step-by-step explanation:

2x + 5 = 12

-- minues five from both sides --

2x = 7

-- divide both sides by two ---

x = 3.5

<em><u>Please give brainliest. </u></em>

4 0

4 years ago

Each student in fifth grade at Blanton elementary brought $15 to the book fair.they brought $1,245 all together,how many fifth g

dangina [55]

First add 15$ then add $1,245 all together as it says and it should give u the answer

5 0

3 years ago

Read 2 more answers

. The estimate regression equation for a model involving two independent variables and 10 observations follows.y = 29.1270 + 0.5

beks73 [17]

Answer:

b) y = 289.815 when $x_1 = 180 \text{ and } x_2 = 310$

Step-by-step explanation:

We are given the following information in the question:

$y = 29.1270 + 0.5906x_1 + 0.4980x_2$

where y is the dependent variable,

$x_1, x_2$ are the independent variable.

The multiple regression equation is of the form:

$y = b_0 + b_1x_1 + b_2x_2$

where,

$b_0$ : is the intercept of the equation and is the value of dependent variable when all the independent variable are zero.

$b_1$ : It is the slope coefficient of the independent variable $x_1$ .

$b_2$ : It is the slope coefficient of the independent variable $x_2$ .

The regression coefficient in multiple regression is the slope of the linear relationship between the dependent and the part of a predictor variable that is independent of all other predictor variables.

Comparing the equations, we get:

$b_1 = 0.5906\\b_2 = 0.4980$

This means holding $x_2$ constant, a change of one in $x_1$ is associated with a change of 0.5906 in the dependent variable.

This means holding $x_1$ constant, a change of 1 in $x_2$ is associated with a change of 0.4980 in the dependent variable.

b) We have to estimate the value of y

$y = 29.1270 + 0.5906(180) + 0.4980(310) = 289.815$

3 0

3 years ago

If Ben invests $4500 at 4% interest per year, how much additional money must he invest at 5 1/2% annual interest to ensure that

pogonyaev

$Step \; 1: \; Assing \; variable \; for \; the \; unknown \; that \; we \; need \; to \; find.\\Let \; x \; be \; additional \; money \; invested$

$Step \; 2: \; Use \; the \; appropriate \; formula \; of \; simple \; interest \; set \; up \; an \; equation\\Interest \; earned \; each \; year = \; Amount \; invested \times annual \; rate \; of \; interest\\\\Amount \; invested \; in \; 4 \% \; rate \; of \; interest=\$4500\\Amount \; invested \; in \; 5\frac{1}{2}\% \; rate \; of \; interest= \$x\\Amount \; invested \; totally \; in \; 4\frac{1}{2}\% \; rate \; of \; interest = \$(4500+x)$

$Interest \; earned \; in \; 4\% \; interest \; rate\\ = \; 4500(4\%)=180\\Interest \; earned \; in \; 5\frac{1}{2}\% \; interest \; rate\\ = x(5\frac{1}{2}\%)=0.055x\\Interest \; earned \; totally \; in \; 4\frac{1}{2}\% \; interest \; rate = (4500+x)(4\frac{1}{2}\%)\\=0.045(4500+x)$

$\\\\So, \; the \; equation \; would \; be:\\180+0.055x=0.045(4500+x)$

$Step \; 1: \; Distribute \; 0.045 \; in \; the \; right \; side\\180+0.055x=202.5+0.045x\\\\Step \; 2: \; Subtract \; 0.045x \; and \; 180 \; on \; both \; sides \; to \; get \; x \; alone\\0.055x-0.045x=202.5-180\\\\Step \; 3: \; Combine \; Like \; Terms\\0.01x=22.5\\\\Step \; 4: Divide \; 0.01 \; on \; both \; sides\\\frac{0.01x}{0.01}=\frac{22.5}{0.01}\\\\Step \; 5: \; Simplifying \; fraction \; on \; both \; sides\\x=2250$

$\underline{Conclusion:}\\Ben \; should \; invest \; additional \; money \; of \; \$2250$

5 0

4 years ago