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vekshin1
3 years ago
5

Angela is putting a fence around her garden, which is bordered by her house on one side (as shown in the picture).

Mathematics
1 answer:
joja [24]3 years ago
5 0

Answer:

B:5x

Step-by-step explanation:

It would be B because if you add all the sides together you would get 5x and 5x is B

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What are the zeros of the quadratic function represented by this graph?​
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Answer:

Step-by-step explanation:

answer : C  when this graph passes trough x-axis

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The coordinates of the vertices of ANGLE RST are R(-3,5), and S(4,5), and T(4,-2). Find the side lengths to the nearest hundredt
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Side lengths: RS=7 and ST=7, and angle=90 degrees

Why?

Since second coordinates of R and S are the same so we can just count the length by adding first coordinate of R and first coordinate of S= |-3|+4=7

Since first coordinates of R is the same as first coordinate of T so we can just count  the length  by adding second coordinates of S and T=5+|-2|=7

Angle: RST is =90 degrees because triangle RST is right angled  triangle. Why? Because RS is parallel to X axis(the same second coordinates of R and S) and ST is parallel to Y axis(the same coordinates of S and T) .


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3 years ago
HELP ASAP!!! DUE SOON!!!<br> 7 7/8 - 3 9/16​
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Answer:

4 5/16

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7 7/8 - 3 9/16

7 14/16 - 3 9/16

4 5/16

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3 years ago
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If you need $8000 to travel to Europe after you graduate in 4 year. How much would your monthly deposits need to be if your acco
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\bf \qquad \qquad \textit{Future Value of an ordinary annuity}\\&#10;\left. \qquad \qquad \right.(\textit{payments at the end of the period})&#10;\\\\&#10;A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]

\bf \qquad &#10;\begin{cases}&#10;A=&#10;\begin{array}{llll}&#10;\textit{accumulated amount}\\&#10;\end{array}\to &&#10;\begin{array}{llll}&#10;8000&#10;\end{array}\\&#10;pymnt=\textit{periodic payments}\\&#10;r=rate\to 5\%\to \frac{5}{100}\to &0.05\\&#10;n=&#10;\begin{array}{llll}&#10;\textit{times it compounds per year}\\&#10;\textit{monthly, thus twelve}&#10;\end{array}\to &12\\&#10;t=years\to &4&#10;\end{cases}

\bf 8000=pymnt\left[ \cfrac{\left( 1+\frac{0.05}{12} \right)^{12\cdot 4}-1}{\frac{0.05}{12}} \right]&#10;\\\\\\&#10;\cfrac{8000}{\left[ \frac{\left( 1+\frac{0.05}{12} \right)^{12\cdot 4}-1}{\frac{0.05}{12}} \right]}=pymnt\implies \cfrac{8000}{\frac{\left( \frac{241}{240} \right)^{48}-1}{\frac{1}{240}}}=pymnt&#10;\\\\\\&#10;\cfrac{8000}{53.0148852}\approx pymnt\implies 150.9010152318\approx pymnt
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3 years ago
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