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3 years ago

Which equation has the solutions x=5+-2 square root7/3

Mathematics

1 answer:

Iteru [2.4K]3 years ago

5 0

Answer:

$3x^2 -10x - 1 = 0$

Step-by-step explanation:

We need to find an equation that has the solutions x=(5±2√7)/3

We know that for a polynomial of the type ax^2 + bx + c = 0, the solution to that polynomial is given by the quadratic formula:

$\frac{-b+\sqrt{b^2-4ac} }{2a}$

In this case, we know that:

-b = 5 → b = -5

2a = 3 → a=3/2

b^2-4ac = 28 → (-5)^2-4(3/2)c = 28 → 25 - 6c = 28 → -6c = 3 → c= -1/2.

Therefore, the equation is:

$\frac{3}{2} x^2 -5x -\frac{1}{2} = 0$

$3x^2 -10x - 1 = 0$

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2x + y = 2 -3x + y = 4 Transfer these into y=mx+b

dimaraw [331]

Step-by-step explanation:

2x + y = 2 subtract 2x from both sides

2x - 2x + y = 2 - 2x

y = -2x + 2

(m = -2, b = 2)

-3x + y = 4 add 3x to both sides

-3x + 3x + y = 4 + 3x

y = 3x + 4

(m = 3, b = 4)

7 0

4 years ago

The sample space of picking a 2

Alchen [17]

Answer:

$Pr = \frac{1}{15}$

Step-by-step explanation:

Given

See attachment for sample space

Required

$P(1\ Y\ and\ 1\ P)$

From the attached sample space, we have:

$n(S) =30$ ---- total outcomes

$n(PY) = 2$ --- outcomes with 1 Y and 1 P = 2 i.e. (P,Y) and (Y,P)

So, the required probability is:

$Pr = \frac{n(PY)}{n(S)}$

$Pr = \frac{2}{30}$

$Pr = \frac{1}{15}$

The error is that; Carlos omitted the numerator

4 0

3 years ago

The cost of 300 metres optic fibre is $200 what will be the cost if 1 metre of optic fibre.

Elden [556K]

The answer is: " $0.67 / metre " ;
or, write as: " [ $\frac{2}{3}$ dollar] per metre."
____________________

Step-by-step explanation:

This is a unit rate problem:
Find the cost in dollars per [single unit—in this case: "metre(s)", "m" ;
_________________
$\frac{200 dollars}{300m} = \frac{? dollars}{1m}$ ; Solve for the "?" amount of dollars.
_________________
Method 1) "Cross-factor multiply" :
Note: Given: " $\frac{a}{b} = \frac{c}{d} ; [b\neq 0; d\neq 0]$ " ;

then: ⟷ " $ad = bc$ " ;
_________________

So: " \frac{200 dollars}{300m} = \frac{? dollars}{1m} " ;

then: ⟷ " (200*1) = (300* ?) ;

→ " (200) = (300 * ?)" ;

↔ " (300 * ?) = 200 " ;

Let "x" refer to the "?" ; the "unknown value" (in dollars);

for which we are trying to solve.

⇒ " 300x = 200 " ; Solve for "x" ;
→ Divide each side of the equation by: "300" ;

to isolate "x" on the "left-hand side" of the equation; & to solve for "x" ;
⇒ 300x/ 300 = 200/300 ;

To get: " x = $\frac{2}{3}$ dollars" ;

= $0.666666666... ;

round to: $0.67 .

{since: " 1 dollar = $1.00 = 100 cents" ;
and since: "\frac{2}{3} " ;

= ["2 ÷ 3" = 0.666666...." ] ;

⇒ round to: "0.67 " ;

and write as:
__________

⇒ $0.67 / metre ; or, write as:
" [ $\frac{2}{3}$ dollar] per metre."
_________

Method 2) Note:
\frac{200 dollars}{300m} = \frac{? dollars}{1m} ;

⇔ = " $\frac{200}{300}$ $\frac{dollars}{metre}$ " ;
{Note: "[200÷300]" can be simplified by "canceling out" the two (2) zeros in both the numerator and the denominator.}.

= " \frac{2 dollars}{3m} ";
_________

= i.e. " $\frac{\frac{2}{3} dollars}{metre}$ " or; $\frac{0.67 dollars} {m}$ ; or: $0.67 / metre .

⇒ {since: " $\frac{2}{3}$ of 1 dollar" = " \frac{2}{3} dollars of 100 cents" ;
⇒ {since: "1 dollar = 100 cents"} ;

⇒ {and: " $\frac{2}{3}$ " = 0.6666666666666... " } ;

→ round to: "0.67 " ; and thus: $0.67 / metre.
_________
Note that using either of these 2 (two) methods result in the same answer.
Hope this is helpful! Best wishes to you within your academic pursuits!
_________________

7 0

3 years ago

-18 divided by -32 rational

jasenka [17]

The answer would be 9/16

5 0

3 years ago

algerbra. com The ninth graders at a high school are raising money by selling T-shirts and baseball caps. The number of T-shirts

My name is Ann [436]

Answer:

$36$ T-shirts and $12$ Baseball Caps

Step-by-step explanation:

GIVEN: The ninth graders at a high school are raising money by selling T-shirts and baseball caps. The number of T-shirts sold was three times the number of caps. The profit they received for each T-shirt sold was $\$5.00$ , and the profit on each cap was $\$2.50$ .