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3 years ago

Find the value of X I need help it’s in math class

Mathematics

2 answers:

Eddi Din [679]3 years ago

6 0

The answer is twelve :)

eduard3 years ago

4 0

Answer:

x = 6

Step-by-step explanation:

The midsegment is half the measure of the parallel side 36, thus

3x = 18 ( divide both sides by 3 )

x = 6

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Suppose the roots of the equation 2x^2−5x−6=0 are α and β.Find the quadratic equation with roots 1/α and 1/β.

pochemuha

Answer:

6x² + 5x - 2 = 0

Step-by-step explanation:

Given

2x² - 5x - 6 = 0 ← in standard form

with a = 2, b = - 5, c = - 6, then

sum of roots α + β = - $\frac{b}{a}$ = $\frac{5}{2}$

product of roots = $\frac{c}{a}$ = - 3, then

sum of new roots = $\frac{1}{\alpha }$ + $\frac{1}{\beta }$

= $\frac{\beta+\alpha }{\alpha \beta }$

= $\frac{\frac{5}{2} }{-3}$ = - $\frac{5}{6}$

product of new roots = $\frac{1}{\alpha }$ × $\frac{1}{\beta }$

= $\frac{1}{\alpha\beta }$ = - $\frac{1}{3}$

Hence the required equation is

x² + $\frac{5}{6}$ x - $\frac{1}{3}$ = 0 or

6x² + 5x - 2 = 0 ( multiplying through by 6 )

8 0

3 years ago

Solve the following using Substitution method<br> 2x – 5y = -13<br><br> 3x + 4y = 15

Digiron [165]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Solve the following using Substitution method

2x – 5y = -13

3x + 4y = 15

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\left. \begin{array} { l } { 2 x - 5 y = - 13 } \\ { 3 x + 4 y = 15 } \end{array} \right.$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$2x-5y=-13, \: 3x+4y=15$

Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.

$2x-5y=-13$

Add 5y to both sides of the equation.

$2x=5y-13$

Divide both sides by 2.

$x=\frac{1}{2}\left(5y-13\right) \\$

Multiply $\frac{1}{2}\\$ times 5y - 13.

$x=\frac{5}{2}y-\frac{13}{2} \\$

Substitute $\frac{5y-13}{2}\\$ for x in the other equation, 3x + 4y = 15.

$3\left(\frac{5}{2}y-\frac{13}{2}\right)+4y=15 \\$

Multiply 3 times $\frac{5y-13}{2}\\$ .

$\frac{15}{2}y-\frac{39}{2}+4y=15 \\$

Add $\frac{15y}{2} \\$ to 4y.

$\frac{23}{2}y-\frac{39}{2}=15 \\$

Add $\frac{39}{2}\\$ to both sides of the equation.

$\frac{23}{2}y=\frac{69}{2} \\$

Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.

$\large \underline{ \underline{ \sf \: y=3 }}$

Substitute 3 for y in $x=\frac{5}{2}y-\frac{13}{2}\\$ . Because the resulting equation contains only one variable, you can solve for x directly.

$x=\frac{5}{2}\times 3-\frac{13}{2} \\$

Multiply 5/2 times 3.

$x=\frac{15-13}{2} \\$

Add $-\frac{13}{2}\\$ to $\frac{15}{2}\\$ by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.