The solutions of the quadratic equation x² - 5x - 36 = 0 are -4 and 9.
The graph is attached below.
- We are given a quadratic function.
- A polynomial equation of degree two in one variable is a quadratic equation.
- The function given to us is :
- y = x² - 5x - 36
- We need to find the solution of the quadratic function.
- To find the roots, let y = 0.
- x² - 5x - 36 = 0
- Use the quadratic formula.
- In elementary algebra, the quadratic formula is a formula that gives the solution(s) to a quadratic equation.
- x = [-b±√b²-4ac]/2a
- x = [-(-5) ± √25 - 4(1)(-36)]/2(1)
- x = (5 ± √25 + 144)/2
- x = (5 ± √169)/2
- x = (5 ± 13)/2
- x = 9 or x = -4
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Answer:
Cube has larger volume than cylinder.
Step-by-step explanation:
Given:
Side of a cube = 7 cm
Radius of a cylinder (r) = 5 cm
Height of a (h) = 4 cm
Find:
Larger volume = ?
Computation:
Volume of cube = Side³
Volume of cube = 7³
Volume of cube = 343 cm³
Cube has larger volume than cylinder.
Answer:
a. H0:μ1≥μ2
Ha:μ1<μ2
b. t=-3.076
c. Rejection region=[tcalculated<−1.717]
Reject H0
Step-by-step explanation:
a)
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.
Answer:
The third one down 0r abc is congruent to edc because m<3=m<4&m<
Step-by-step explanation:
this is because you need three point to prove the congruency of triangles
