2x + 5y = -3 ⇒ 2x + 5y = -3
1x + 8y = 4 ⇒ <u>2x + 16y = 8
</u> -<u>11y</u> = <u>-11 </u>
-11 -11
y = 1
2x + 5(1) = -3
2x + 5 = -3
<u> -5 -5</u>
<u>2x</u> = <u>-8</u>
2 2
x = -4
(x, y) = (-4, 1)
2x + 1y = 7 ⇒ 2x + 1y = 7
1x - 2y = -14 ⇒ <u>2x - 4y = -28</u>
<u>5y</u> = <u>35</u>
5 5
y = 7
2x + 7 = 7
<u> -7 -7</u>
<u>2x</u> = <u>0</u>
2 2
x = 0
(x, y) = (0, 7)
Answer:
![A_T=150cm^2](https://tex.z-dn.net/?f=A_T%3D150cm%5E2)
Explanation: We need to find the area of the triangle, We know that the area of any right triangle is:
![A_T=\frac{1}{2}(b\times h)](https://tex.z-dn.net/?f=A_T%3D%5Cfrac%7B1%7D%7B2%7D%28b%5Ctimes%20h%29)
Therefore in the information given we have the following:
![\begin{gathered} h=20\operatorname{cm} \\ b=15\operatorname{cm} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20h%3D20%5Coperatorname%7Bcm%7D%20%5C%5C%20b%3D15%5Coperatorname%7Bcm%7D%20%5Cend%7Bgathered%7D)
Therefore the Area of this triangle is as follows:
![\begin{gathered} A_T=\frac{1}{2}(b\times h) \\ \therefore\rightarrow \\ A_T=\frac{1}{2}(20cm\times15cm)=\frac{1}{2}(20\times15)cm^2 \\ A_T=\frac{1}{2}(300)cm^2=150cm^2 \\ \therefore\rightarrow \\ A_T=150cm^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A_T%3D%5Cfrac%7B1%7D%7B2%7D%28b%5Ctimes%20h%29%20%5C%5C%20%5Ctherefore%5Crightarrow%20%5C%5C%20A_T%3D%5Cfrac%7B1%7D%7B2%7D%2820cm%5Ctimes15cm%29%3D%5Cfrac%7B1%7D%7B2%7D%2820%5Ctimes15%29cm%5E2%20%5C%5C%20A_T%3D%5Cfrac%7B1%7D%7B2%7D%28300%29cm%5E2%3D150cm%5E2%20%5C%5C%20%5Ctherefore%5Crightarrow%20%5C%5C%20A_T%3D150cm%5E2%20%5Cend%7Bgathered%7D)
Note! Units are centimeter-squared
he elements of the Klein <span>44</span>-group sitting inside <span><span>A4</span><span>A4</span></span> are precisely the identity, and all elements of <span><span>A4</span><span>A4</span></span>of the form <span><span>(ij)(kℓ)</span><span>(ij)(kℓ)</span></span> (the product of two disjoint transpositions).
Since conjugation in <span><span>Sn</span><span>Sn</span></span> (and therefore in <span><span>An</span><span>An</span></span>) does not change the cycle structure, it follows that this subgroup is a union of conjugacy classes, and therefore is normal.
Answer:
60 or 60:1
Step-by-step explanation:
You will have to divide 180 by 3 which is 60 or you can divide 160 / 3 = 60 & divide 3 / 3 = 1.
Those are the ranges for one tandard deviation below the mean and one standard deviation above the mean, and we know that 68% of the data falls in that range, so the probability that a randomly chosen worker makes that much is 68%