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2 years ago

I have no idea what to do with this!

Mathematics

2 answers:

lyudmila [28]2 years ago

4 0

Answer:

guy above me is right ✅

finlep [7]2 years ago

3 0

The function given in the question is

<span>g(x) = 4x - 5
When x = 2, then
g(2) = (4 * 2) - 5
= 8 -5
= 3
When x = - 4, then
g(-4) = (4 * -4) - 5
= - 16 - 5
= - 21
From the above deduction, it can be concluded that the correct option among all the options given in the question is the third option or "</span><span>The value of g(2) is larger than the value of g(-4)". I hope that the procedure is clear enough for you to understand.</span>

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X-5(x-1)=2x-(2x-3) answer the question

levacccp [35]

Answer: x= 1/2

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

x−5(x−1)=2x−(2x−3)

x−5(x−1)=2x+−1(2x−3)(Distribute the Negative Sign)

x−5(x−1)=2x+−1(2x)+(−1)(−3)

x−5(x−1)=2x+−2x+3

x+(−5)(x)+(−5)(−1)=2x+−2x+3(Distribute)

x+−5x+5=2x+−2x+3

(x+−5x)+(5)=(2x+−2x)+(3)(Combine Like Terms)

−4x+5=3

Step 2: Subtract 5 from both sides.

−4x+5−5=3−5

−4x=−2

Step 3: Divide both sides by -4.

−4x /−4 = −2 /−4

x = 1/2

8 0

3 years ago

Read 2 more answers

2.1 + 3.32 – 1.4 = ? A. 6.82 B. 4.02 C. 5.42 D. 4.2

goblinko [34]

4.02 is the answer., B.

8 0

3 years ago

Read 2 more answers

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

Someone Explain to me how 5. Is valid ???

vichka [17]

This is valid through the law of syllogism. If you swap lines 1 and 2, then you'll have this argument:

If I step on a beehive, then I will get stung.

If I get stung by a bee, then it will hurt.

Therefore, if I step on a beehive, then it will hurt

------------------

So it's like connecting a chain together. Point A (stepping on the hive), leads to point B (getting stung), which leads to point C (getting hurt). We can take a shortcut to bypass point B to jump from A to C in one step. Check out the attached image for a visual of what I'm referring to.

6 0

3 years ago

SAT scores are distributed with a mean of 1,500 and a standard deviation of 300. You are interested in estimating the average SA

kotegsom [21]

Answer:

I should use at least 304 students

Step-by-step explanation:

Margin error (E) = t × sd/√n

E = 40

sd = 300

confidence level (C) = 98% = 0.98

significance level = 1 - C = 1 - 0.98 = 0.02 = 2%

t-value corresponding to 2% significance level and infinity degree of freedom is 2.326

n = (t×sd/E)^2 = (2.326×300/40)^2 = 17.445^2 = 304 (to the nearest integer)

3 0

3 years ago