Answer:
Use the distance formula on both points AC and AB.
<em>Distance formula is this</em><em>:</em>
<em>\begin{gathered}d=\sqrt{(x2-x1)^2+(y2-y1)^2} \\\\d=\sqrt{(1--5)^2+(8--7)^2} \\\\d=\sqrt{(6)^2+(15)^2} \\\\d=\sqrt{36+225} \\\\d=\sqrt{261} \\\\\end{gathered}d=(x2−x1)2+(y2−y1)2d=(1−−5)2+(8−−7)2d=(6)2+(15)2d=36+225d=261</em>
Distance for AC is 16.16
Now do the same with the numbers for AB and get the distance of 5.39
2. To get the area, use the formula 1/2 x base x height
AB is the base and AC is the height.
1/2 x 16.16 x 5.39 = 43.55
the answer is 43.5
Answer:
RC = 40
Step-by-step explanation:
Note that the circumcentre is equally distant from the triangle's 3 vertices.
That is : PC = RC = QC
Equate any pair and solve for x
Using RC = PC, then
5x - 15 = 3x + 7 ( subtract 3x from both sides )
2x - 15 = 7 ( add 15 to both sides )
2x = 22 ( divide both sides by 2 )
x = 11
Hence
RC = (5 × 11) - 15 = 55 - 15 = 40 units
Answer:
c.
Step-by-step explanation:
The following is how the p-value should be interpreted:
The probability of getting the test statistics value at least as intense as it might have been acquired during the test, assuming the null hypothesis is true, is considered as the p-value.
Null hypothesis
A student's grade level has no bearing or association on how they commute to school.
Alternative hypothesis;
There is a link between a student's grade level and the way they get to school.
The test statistic's value is 14.63, and the p-value is 0.2636.
As a result, the p-value is interpreted as that of option c in the question.
F(x)= (x-p)²+q where (p,q) is min point
f(x)=(x-1)²-2
min point is (1,-2)
the answer is first picture
