-3/4 = -21/28
1/14 = 2/28
-21 + 2 = -19
the answer would be:
-19/28
Hope this helps!
Answer:
D
Step-by-step explanation:
<u>QUESTION 1</u>
The given figure is made up of a parallelogram and a rectangle.
The area of the parallelogram is
![=base\times height](https://tex.z-dn.net/?f=%3Dbase%5Ctimes%20height)
The base of the parallelogram is 6 units (Just count the boxes).
The height is also 1 unit.
This implies that the area of the parallelogram is
![=6\times 1 \:units^2](https://tex.z-dn.net/?f=%3D6%5Ctimes%201%20%5C%3Aunits%5E2)
![=6\:units^2](https://tex.z-dn.net/?f=%3D6%5C%3Aunits%5E2)
To find the area of the rectangle, we need to find the width and the length using the distance formula or the Pythagoras Theorem.
Using the Pythagoras Theorem,
![l^2=a^2+b^2](https://tex.z-dn.net/?f=l%5E2%3Da%5E2%2Bb%5E2)
![\Rightarrow l^2=2^2+6^2](https://tex.z-dn.net/?f=%5CRightarrow%20l%5E2%3D2%5E2%2B6%5E2)
![\Rightarrow l^2=4+36](https://tex.z-dn.net/?f=%5CRightarrow%20l%5E2%3D4%2B36)
![\Rightarrow l^2=40](https://tex.z-dn.net/?f=%5CRightarrow%20l%5E2%3D40)
![\Rightarrow l=\sqrt{40}](https://tex.z-dn.net/?f=%5CRightarrow%20l%3D%5Csqrt%7B40%7D)
![\Rightarrow l=2\sqrt{10}](https://tex.z-dn.net/?f=%5CRightarrow%20l%3D2%5Csqrt%7B10%7D)
Similarly,
![w^2=1^2+3^2](https://tex.z-dn.net/?f=w%5E2%3D1%5E2%2B3%5E2)
![\Rightarrow w^2=1+9](https://tex.z-dn.net/?f=%5CRightarrow%20w%5E2%3D1%2B9)
![\Rightarrow w^2=10](https://tex.z-dn.net/?f=%5CRightarrow%20w%5E2%3D10)
![\Rightarrow w=\sqrt{10}](https://tex.z-dn.net/?f=%5CRightarrow%20w%3D%5Csqrt%7B10%7D)
The area of the rectangle is
![Area=l\times w](https://tex.z-dn.net/?f=Area%3Dl%5Ctimes%20w)
We substitute the values into the formula to obtain;
![Area=2\sqrt{10}\times \sqrt{10}](https://tex.z-dn.net/?f=Area%3D2%5Csqrt%7B10%7D%5Ctimes%20%5Csqrt%7B10%7D)
![Area=2\times10](https://tex.z-dn.net/?f=Area%3D2%5Ctimes10)
![Area=20\:units^2](https://tex.z-dn.net/?f=Area%3D20%5C%3Aunits%5E2)
The area of the figure is
![=20+6](https://tex.z-dn.net/?f=%3D20%2B6)
![=26\:units^2](https://tex.z-dn.net/?f=%3D26%5C%3Aunits%5E2)
<u>QUESTION 2</u>
We can divide the given polygon into two parts to obtain a triangle and a rectangle. See diagram in attachment.
The area of the triangular portion is
![=\frac{1}{2}\times base\times height.](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20base%5Ctimes%20height.)
![=\frac{1}{2}\times 9\times 6.](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%209%5Ctimes%206.)
![=9\times 3.](https://tex.z-dn.net/?f=%3D9%5Ctimes%203.)
![=27\:units^2.](https://tex.z-dn.net/?f=%3D27%5C%3Aunits%5E2.)
The area of the rectangle is
![=l\times w](https://tex.z-dn.net/?f=%3Dl%5Ctimes%20w)
![=9\times 2](https://tex.z-dn.net/?f=%3D9%5Ctimes%202)
![=18\:units^2](https://tex.z-dn.net/?f=%3D18%5C%3Aunits%5E2)
The area of the polygon is
![=27+18\:units^2](https://tex.z-dn.net/?f=%3D27%2B18%5C%3Aunits%5E2)
![=45\:units^2](https://tex.z-dn.net/?f=%3D45%5C%3Aunits%5E2)
V=V of cube- V pyramid= 8^3- 7^2*8/3=512-49*8/3=512-130,67=381.33 cm^3
First expression simplified would be
-2z - 63
second expression would be
-a + b
i hope that’s right!!!!!!! i’m so sorry if it’s not