When you have something like this, all you need to do is substitute the values, the last is for what value of x
For the first one;
((x^2+1)+(x-2))(2)
(x^2+x-1)(2)
(2)^2+(2)-1
4+2-1
5
For the second one;
((x^2+1)-(x-2))(3)
(x^2-x+3)(3)
(3)^2-(3)+3
9-3+3
9
For the last one;
3(x^2+1)(7)+2(x-2)(3)
3((7)^2+7)+2((3)-2)
3(49+7)+2(3-2)
3(56)+2(1)
168+2
170
<span>The missing angle measure in triangle ABC is 55°.
The measure of angle BAC in triangle ABC is equal to the measure of angle
EDF in triangle DEF.
The measure of angle ABC in triangle ABC is equal to the measure of </span><span>angle EFD in triangle DEF.
Triangles ABC and DEF are similar by the angle-angle criterion.
True </span>
Step-by-step explanation:
Given that the graph shows the normal distribution of the length of similar components produced by a company with a mean of 5 centimeters and a standard deviation of 0.02 centimeters.
A component is chosen at random, the probability that the length of this component is between 4.98 centimeters and 5.02
=P(|z|<1) (since 1 std dev on either side of the mean)
=2(0.3418)
=0.6826
=68.26%
The probability that the length of this component is between 5.02 centimeters and 5.04 centimeters is
=P(1<z<2) (since between 1 and 2 std dev from the mean)
=0.475-0.3418
=0.3332
=33.32%
Yes you do and what grade are you in and do you need help answering these questions
