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1 year ago

What’s the answer for this equation: x<-x+7(x-2)

Mathematics

1 answer:

klio [65]1 year ago

8 0

Answer:

x > 2.8

Step-by-step explanation:

x < -x + 7(x - 2)

Expand using the Distributive Property

x < -x + 7x - 14

x < 6x - 14

Subtract x from both sides

0 < 6x - 14 - x

0 < 5x - 14

Add 14 for both sides

5x > 14

x > 14/5

x > 2.8

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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

5^(-x)+7=2x+4 This was on plato

Setler79 [48]

Answer:

Below

I hope its not too complicated

$x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

Step-by-step explanation:

$5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}$

$Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}$

$\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x$

$\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}$

3 0

3 years ago

Find the domain of h(x)=x-3/x^3-36x

Dovator [93]

There is some ambiguity here which could be removed by using parentheses. I'm going to assume that you actually meant:

x-3

h(x) = ---------------

(x^3-36x)

To determine the domain of this function, factor the denominator:

x^3 - 36x = x(x^2 - 36) = x(x-6)(x+6)

The given function h(x) is undefined when the denominator = 0, which happens at {-6, 0, 6}.

Thus, the domain is "the set of all real numbers not equal to -6, 0 or 6."

Symbolically, the domain is (-infinity, -6) ∪ (-6, 0) ∪ (0, 6) ∪ (6, +infinity).

3 0

3 years ago

Expand. Your answer should be a polynomial in standard form. (b^2+5)(−b^2+7)=

Blizzard [7]

(b^2+5)(−b^2+7)

Use the FOIL METHOD to expand.

-b^4 + 7b^2 - 5b^2 + 35

-b^4 + 2b^2 + 35

Did you follow?

6 0

2 years ago

A proportion often shows the relationship between:

tankabanditka [31]

Answer:

It compares two numbers that can represent many of things. For example, people, food & items.

Step-by-step explanation:

8 0

2 years ago