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3 years ago

What is the input value for which the statement f(x) = g(x) is true?

Mathematics

1 answer:

Murljashka [212]3 years ago

4 0

Answer: According with the graph, the input value for which the statement f(x)=g(x) is true (the value of "x" where the graph intersect) is 1.5 (third option).

Solution

The input value for which the statement f(x) = g(x) is true, is the value of "x" where the graph of the two functions intersect. Accordind with the graph, the two functions f(x) and g(x) intersect at x between 1 and 2, approximately at x=1.5

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Please help... I have no clue

muminat

Answer:

OPTION C: Sin C - Cos C = s - r

Step-by-step explanation:

ABC is a right angled triangle. ∠A = 90°, from the figure.

Therefore, BC = hypotenuse, say h

Now, we find the length of AB and AC.

We know that: $\textbf{Sin A} = \frac{\textbf{opp}}{\textbf{hyp}}$

and $\textbf{Cos A} = \frac{\textbf{adj}}{\textbf{hyp}}$

Given, Sin B = r and Cos B = s

⇒ $Sin B = r = \frac{opp}{hyp} = \frac{AC}{BC} = \frac{AC}{h}$

⇒ $\textbf{AC} = \textbf{rh}$

Hence, the length of the side AC = rh

Now, to compute the length of AB, we use Cos B.

$Cos B = s = \frac{adj}{hyp} = \frac{AB}{BC} = \frac{AB}{h}$

⇒ $\textbf{AB} = \textbf{sh}$

Hence, the length of the side AB = sh

Now, we are asked to compute Sin C - Cos C.

$Sin C = \frac{opp}{hyp}$

⇒ $Sin C = \frac{AB}{BC}$

$= \frac{sh}{h}$

= s

Sin C = s

$Cos C = \frac{adj}{hyp}$

$\implies Cos C = \frac{AC}{BC}$

⇒ Cos C = $\frac{rh}{h}$

Therefore, Cos C = r

So, Sin C - Cos C = s - r, which is OPTION C and is the right answer.

5 0

3 years ago

What are the answers to the last two questions?

ivanzaharov [21]

The answer to 15 is addition and multiplication

the answer to 16 is 0

8 0

3 years ago

Solve the following equation. Remember to check for extraneous solutions 1/(x-6) + (x/(x-2)) = (4/(x²-8x+12)).

xxMikexx [17]

Answer:

-1, 2, 6

Step-by-step explanation:

We have to solve the equation as follows: 1/(x-6) + (x/(x-2)) = (4/(x²-8x+12)).

Now, we have, $\frac{1}{x-6} +\frac{x}{x-2} = \frac{4}{x^{2}-8x+12 }$

⇒ $\frac{(x-2)+x(x-6)}{(x-2)(x-6)} = \frac{4}{x^{2}-8x+12 }$

⇒ $\frac{x-2+x^{2}-6x }{(x-2)(x-6)} =\frac{4}{(x-2)(x-6)}$

⇒ $\frac{(x-2)(x-6)}{x^{2}-5x-2 }=\frac{(x-2)(x-6)}{4}$

⇒ $(x-2)(x-6)[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0$

⇒ $(x-2)(x-6) =0$ or, $[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0$

If, (x-2)(x-6) =0, then x=2 or x=6

If, $[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0$ , then $x^{2} -5x-2=4$

and (x-6)(x+1) =0

Therefore, x=6 or -1

So the solutions for x are -1, 2 6. (Answer)

4 0

3 years ago

The following 5 questions are based on this information: An economist claims that average weekly food expenditure of households

liq [111]

Answer:

Null hypothesis: $\mu_{1}-\mu_{0}\leq 0$

Alternative hypothesis: $\mu_{1}-\mu_{2}>0$

$SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705$

b) 2.70

$t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850$

b. 1.85

$p_v =P(Z>1.85)=0.032$

b. 0.03

a. We can reject H(0) in favor of H(a)

Step-by-step explanation:

Data given and notation

$\bar X_{1}=164$ represent the mean for the sample 1

$\bar X_{2}=159$ represent the mean for the sample 2

$\sigma_{1}=12.5$ represent the population standard deviation for the sample 1

$s_{2}=9.25$ represent the population standard deviation for the sample B2

$n_{1}=35$ sample size selected 1

$n_{2}=30$ sample size selected 2

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the expenditure of households in City 1 is more than that of households in City 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{0}\leq 0$

Alternative hypothesis: $\mu_{1}-\mu_{2}>0$

We know the population deviations, so for this case is better apply a z test to compare means, and the statistic is given by:

$z=\frac{(\bar X_{1}-\bar X_{2})-0}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Standard error

The standard error on this case is given by:

$SE=\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}$

Replacing the values that we have we got:

$SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705$

b. 2.70

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850$

P-value

Since is a one side right tailed test the p value would be:

$p_v =P(Z>1.85)=0.032$

b. 0.03

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

a. We can reject H(0) in favor of H(a)

4 0

3 years ago

What must be a factor of the polynomial function f(x) graphed on the coordinate plane below?

Nady [450]

The correct answer is x because u have to look on the x axis

4 0

3 years ago