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3 years ago

If a polynomial function f(x) has roots 3 and StartRoot 7 EndRoot, what must also be a root of f(x)?

Mathematics

2 answers:

irakobra [83]3 years ago

8 0

Answer:

Step-by-step explanation:

Edge 2021

Ivenika [448]3 years ago

4 0

Answer:

–√(7) must also be a root of f(x)

Step-by-step explanation:

Given that the roots of an equation is 3 and √(7). This means that x = 3 or √(7). Therefore, the function is definitely going to be:

f(x) = (x – 3)(x² – 7)

This is because we are dealing with √(7) and what brought about √(7) was x² = 7.

If this is the case, we know that x² = 7 will give

x = ±√(7)

Hence, our roots are x = 3 or √(7) or –√(7). Therefore, –√(7) must also be a root of f(x).

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Stella [2.4K]

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3 years ago

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3 years ago

A study was designed to compare the attitudes of two groups of nursing students towards computers. Group 1 had previously taken

Leviafan [203]

Answer:

a. H0:μ1≥μ2

Ha:μ1<μ2

b. t=-3.076

c. Rejection region=[tcalculated<−1.717]

Reject H0

Step-by-step explanation:

As the score for group 1 is lower than group 2,

Null hypothesis: H0:μ1≥μ2

Alternative hypothesis: H1:μ1<μ2

b) t test statistic for equal variances

t=(xbar1-xbar2)-(μ1-μ2)/sqrt[{1/n1+1/n2}*{((n1-1)s1²+(n2-1)s2²)/n1+n2-2}

t=63.3-70.2/sqrt[{1/11+1/13}*{((11-1)3.7²+(13-1)6.6²)/11+13-2}

t=-6.9/sqrt[{0.091+0.077}{136.9+522.72/22}]

t=-3.076

c. α=0.05, df=22

t(0.05,22)=-1.717

The rejection region is t calculated<t critical value

t<-1.717

We can see that the calculated value of t-statistic falls in rejection region and so we reject the null hypothesis at 5% significance level.

7 0

4 years ago

Let f(x) = 4x – 5 and g(x) = 3x + 7. Find f(x) + g(x) and state its domain.

lesantik [10]

$f(x)=4x-5,\ g(x)=3x+7\\\\f(x)+g(x)=(4x-5)+(3x+7)=4x-5+3x+7\\\\=(4x+3x)+(-5+7)=7x+2\\\\\text{It's a linear function. The domain is the set of all real numbers.}\\\\Answer:\\\\f(x)+g(x)=7x+2\\D=\mathbb{R}$

8 0

4 years ago

Charlie cannot remember how much he financed to buy his car. He does remember that his monthly payment is $200. His add-on inter

Mazyrski [523]

First find the total payments
Total paid
200×30=6,000 (this is the future value)

Second use the formula of the future value of annuity ordinary to find the monthly payment.
The formula is
Fv=pmt [(1+r/k)^(n)-1)÷(r/k)]
We need to solve for pmt
PMT=Fv÷[(1+r/k)^(n)-1)÷(r/k)]
PMT monthly payment?
Fv future value 6000
R interest rate 0.09
K compounded monthly 12
N=kt=12×(30months/12months)=30
PMT=6000÷(((1+0.09÷12)^(30)
−1)÷(0.09÷12))
=179.09 (this is the monthly payment)

Now use the formula of the present value of annuity ordinary to find the amount of his loan.
The formula is
Pv=pmt [(1-(1+r/k)^(-n))÷(r/k)]
Pv present value or the amount of his loan?
PMT monthly payment 179.09
R interest rate 0.09
N 30
K compounded monthly 12
Pv=179.09×((1−(1+0.09÷12)^(
−30))÷(0.09÷12))
=4,795.15

The answer is 4795.15

5 0

3 years ago