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2 years ago

30 POINTS John opens a savings account and decides to deposit $10 each month. Let x represent the number of

Mathematics

1 answer:

tatuchka [14]2 years ago

7 0

Answer:

7.) y=10x

8.) $140 in 14 months, $60 in 6 months

9.) He will have saved $100 in 10 months and $175 in 17.5 months

10.) The rate of change is 10

11.) Knowing the rate of change helps you in answering this problem because it tells you how much money he saves per month.

12.)

0- $0

1- $10

2-$20

3-$30

4-$40

Step-by-step explanation:

7.) The standard form of a linear equation is y=mx+b

The starting balance is 0 so there is no point to adding b. M is the amount of money he deposits each month so that is 10. You plug in the values and you get y=10x (+0)

8.) y=10(14)

$140

y=10(6)

$60

9.) 100=10x

100/10= 10

10 months

175=10x

175/10=17.5

17.5 months

10.) The rate in change is m so 10.

Hope I explained this in simple enough terms. Have a good day!

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7 0

2 years ago

Read 2 more answers

I need help finding the area

aniked [119]

<h3>Given :</h3>

Base of triangle = 7 yd
Height of triangle = 10 yd

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Area of triangle

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We know:-

When base and height of triangle is given we use this formula:

$\bigstar \boxed{ \rm Area \: of \: triangle = \frac{base \times height}{2} }$

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So:-

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{base \times height}{2} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 10}{2} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times2 }{2} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times\cancel2 }{\cancel2} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle = \dfrac{7 \times 5 \times1 }{1} \\$

$\\ \\$

$\dashrightarrow \sf \: Area \: of \: triangle =7 \times 5$

$\\ \\$

$\dashrightarrow \bf \: Area \: of \: triangle =35 {yd}^{2} \\$

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$\therefore \underline{\textsf{ \textbf {\: Area \: of \: triangle = \red{35}}} { \red{\bf{yd} }^{ \red2} }}$

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$\small\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf \small{Formulas\:of\:Areas:-}}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}$ ]