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3 years ago

Rewrite the following quadratic functions in intercept or factored form. Show your work.

Mathematics

2 answers:

puteri [66]3 years ago

4 0

Answer:

$(2x+3)^2$

Step-by-step explanation:

We are given the following the quadratic function and we are to rewrite it in intercept or factored form:

$y = 9 + 12x + 4x^2$

Rearranging the given function to get:

$y = 4x^2+12x+9$

Rewriting $4$ and $4$ in $y = 4x^2+12x+9$ as perfect squares:

$2^2x^2+12+3^2$

Now applying the exponent rule: $a^mb^m=(ab)^m$

$(2x)^2+12x+3^2$

Rewriting it in the form $(a+b)^2=a^2+2ab+b^2$ :

$(2x)^2+2(2x)(3)+3^2$

Here $a=2x$ and $b=3$ . So the factored form is $(2x+3)^2$

Vlad [161]3 years ago

3 0

Answer:

y = (2x + 3)(2x + 3) = (2x + 3)²

Step-by-step explanation:

We are given a quadratic function and we have to write it in factored form.

y = 9 + 12x + 4x²

y = 4x² + 12x + 9

We can break the mid-term in such a way that when they are multiplied, the factors give a product of 36x² and when added, they give a result of 12x, as show below:

y = 4x² + 6x + 6x + 9

Taking 2x common from the first two variables and 3 from the second two

y = 2x(2x + 3) + 3(2x + 3)

Taking 2x+3 common

y = (2x + 3)(2x + 3) = (2x + 3)²

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Answer:

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8 0

3 years ago

there are 4 planets. the first planet has an orbit of 3 years. The second planet has an orbit of 5 years. The third planet has a

Margarita [4]

Find the LCM of the four numbers (3, 5, 8, 12)

3: 3

5: 5

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12: 2 * 2 * 3

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6 0

3 years ago

Give an example of a quadratic equation with non-real solutions.

alexgriva [62]

Answer:

x² + 1 = 0

Step-by-step explanation:

You can form any with discriminant negative (B²-4AC < 0)

An example could be:

x² + 1 = 0

x² = -1 which has no real roots/solutions

3 0

3 years ago

We wish to see if the dial indicating the oven temperature for a certain model of oven is properly calibrated. Four ovens of thi

antoniya [11.8K]

Answer:

a. Standard deviation: 4.082

Standard error: 2.041

b. The 95% confidence interval for the actual temperature is (298.5, 311.5).

Upper bound: 311.5

Lower bound: 298.5

c. Test statistic t=2.45

P-value = 0.092

d. There is no enough evidence to claim that the dial of the oven is not properly calibrated. The actual temperature does not significantly differ from 300 °F.

e. If we use a significance level of 10% (a less rigorous test, in which the null hypothesis is rejected with with less requirements), the conclusion changes and now there is enough evidence to claim that the dial is not properly calibrated.

This happens because now the P-value (0.092) is smaller than the significance level (0.10), given statististical evidence for the claim.

Step-by-step explanation:

The mean and standard deviation of the sample are:

$M=\dfrac{1}{4}\sum_{i=1}^{4}(305+310+300+305)\\\\\\ M=\dfrac{1220}{4}=305$

$s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{4}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{3}\cdot [(305-(305))^2+(310-(305))^2+(300-(305))^2+(305-(305))^2]}\\\\\\ s=\sqrt{\dfrac{1}{3}\cdot [(0)+(25)+(25)+(0)]}\\\\\\ s=\sqrt{\dfrac{50}{3}}=\sqrt{16.667}\\\\\\s=4.082$

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=305.

The sample size is N=4.

When σ is not known, s divided by the square root of N is used as an estimate of σM (standard error):

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{4.082}{\sqrt{4}}=\dfrac{4.082}{2}=2.041$

The degrees of freedom for this sample size are:

$df=n-1=4-1=3$

The t-value for a 95% confidence interval and 3 degrees of freedom is t=3.18.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=3.18 \cdot 2.041=6.5$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 305-6.5=298.5\\\\UL=M+t \cdot s_M = 305+6.5=311.5$

The 95% confidence interval for the actual temperature is (298.5, 311.5).

This is a hypothesis test for the population mean.

The claim is that the actual temperature of the oven when the dial is at 300 °F does not significantly differ from 300 °F.

Then, the null and alternative hypothesis are:

$H_0: \mu=300\\\\H_a:\mu\neq 300$

The significance level is 0.05.

The sample has a size n=4.

The sample mean is M=305.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=4.028.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{4.082}{\sqrt{4}}=2.041$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{305-300}{2.041}=\dfrac{5}{2.041}=2.45$

The degrees of freedom for this sample size are:

$df=n-1=4-1=3$

This test is a two-tailed test, with 3 degrees of freedom and t=2.45, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>2.45)=0.092$

As the P-value (0.092) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the actual temperature of the oven when the dial is at 300 °F does not significantly differ from 300 °F.

If the significance level is 10%, the P-value (0.092) is smaller than the significance level (0.1) and the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the actual temperature of the oven when the dial is at 300 °C does not significantly differ from 300 °C.

5 0

3 years ago