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2 years ago

Find the Slope A. 2/5 B.5/2 C.-2/5 D.-5/2

Mathematics

2 answers:

Fiesta28 [93]2 years ago

5 0

The answer is a bc it’s pointing upwards which makes it positive and the rise is 5 and the run making it 2/5

kirza4 [7]2 years ago

3 0

2/5 because it’s pointing up wards np

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12 - 3/4 (d + 16) = negative 5

Nutka1998 [239]

Answer: d=20/3

Step By Step:

Step 1, Simplify:

12-3/4(d+16)=-5

(-3/4)d+(12-12)=-5

(-3/4)d=-5

Step 2, Multiply each side by 4/(-3)

(4/-3)*(-3/4)d=(4/-3)*-5

d=20/3

8 0

3 years ago

Read 2 more answers

The scale of this map is 1 inch = 50 miles. The ruler measures the distance between Gary, Indiana, and Traverse City, Michigan,

levacccp [35]

Answer:

<em>150 miles</em>

Step-by-step explanation:

Given the scale on the map using the conversion factor

1 inch = 50 miles

We are to find the equivalent miles between the cities in miles if the distace is 3inches

3inches = x

Divide both expressions

1/3 = 50/x

Cross multiply

x = 3 * 50

x = 150

<em>Hence the actual distance between these two cities in miles is 150 miles </em>

6 0

2 years ago

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:-

$\\ \\$

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

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

$\\ \\$

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

$\\ \\$

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

$\\ \\$

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

$\\ \\$

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

$\\ \\$

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

$\\ \\$

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

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