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jeka94
3 years ago
15

Im not sure if anyone can help me but if you can i would really appreciate it cuz im really confused about this so thank you. al

so for the second screenshot where it says select it has 3 options quotient, product, and sum .

Mathematics
1 answer:
Advocard [28]3 years ago
6 0

Answer:

Question 1) 6x+5

Question 2) Blank #1: sum. Blank #2: 5. Blank #3: 4

See work in the attachment. If you're still confused, read the step-by-step explanation for a more in-depth explanation of what exactly I did for my work. Let me know if you have any questions :>

Step-by-step explanation:

For question 1, it's important to know that the perimeter is equal to the sum of all the sides. Since the shape is a rectangle, the sides opposite of each other are the same length. Thus, the left and right sides are both 3x - 3. 2(3x - 3) = 6x - 6. The unknown value, which I called a, is the same for both the top and the bottom. To find it, you must first subtract 6x - 6 from the perimeter of 18x + 4. This equals 12x + 10. Then, divide that answer by 2 to get the unknown side. You then get 6x + 5 as the answer for the unknown side.

To make sure it's correct, you can plug in a random number for x. In this case, I checked my work by plugging 2 in for x.

Perimeter = 18(2) + 4 = 36 + 4 = 40

Perimeter = 6(2) + 5 + 6(2) + 5 + 3(2) - 3 + 3(2) - 3 = 12 + 5 + 12 + 5 + 6 - 3 + 6 - 3 = 24 + 10 + 12 - 6 = 40

For question 2, the main idea you need to know is that A = lw. The area of the first rectangle is 6 * 5 = 30. The area of the second rectangle is 6 * 4 = 24. When you add them together, you get 54. The area of the rectangle that is formed when you combine rectangle 1 and rectangle 2 is 6 * 9 = 54. This shows that 6(4) + 6(5) = 6(9)

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3 years ago
Help me with question a please ! With full workings !
frosja888 [35]
A)


\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
Q&({{ 0}}\quad ,&{{ 2}})\quad 
%  (c,d)
P&({{ 0.5}}\quad ,&{{ 0}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}

\bf QP=\sqrt{(0.5-0)^2+(0-2)^2}\implies QP=\sqrt{0.5^2+2^2}
\\\\\\
QP=\sqrt{\left( \frac{1}{2} \right)^2+4}\implies QP=\sqrt{ \frac{1^2}{2^2}+4}\implies QP=\sqrt{\frac{1}{4}+4}
\\\\\\
QP=\sqrt{\frac{17}{4}}\implies QP=\cfrac{\sqrt{17}}{\sqrt{4}}\implies QP=\cfrac{\sqrt{17}}{2}

b)

since QR=QP, that means that QO is an angle bisector, and thus the segments it makes at the bottom of RO and OP, are also equal, thus RO=OP

thus, since the point P is 0.5 units away from the 0, point R is also 0.5 units away from 0 as well, however, is on the negative side, thus R (-0.5, 0)


c)

what's the equation of a line that passes through the points (-0.5, 0) and (0,2)?

\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%   (a,b)
Q&({{ 0}}\quad ,&{{ 2}})\quad 
%   (c,d)
R&({{ -0.5}}\quad ,&{{ 0}})
\end{array}
\\\\\\
% slope  = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{0-2}{-0.5-0}\implies \cfrac{-2}{-0.5}

\bf m=\cfrac{\frac{-2}{1}}{-\frac{1}{2}}\implies \cfrac{-2}{1}\cdot \cfrac{2}{-1}\implies 4
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-2=4(x-0)\implies y=4x+2\\
\left. \qquad   \right. \uparrow\\
\textit{point-slope form}
7 0
3 years ago
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