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Thepotemich [5.8K]
3 years ago
6

Mr. Vu can drive his new electric car 194 miles before he needs to charge his battery. How far can Mr. Vu’s car go on 15 battery

charges?
A 970 miles
B 1,164 miles
C 1,940 miles
D 2,910 miles
Mathematics
1 answer:
const2013 [10]3 years ago
6 0

Answer:

Mr. Vu can go 2910 miles on 15 battery charges (D).

Step-by-step explanation:

If 1 battery charge lasts 194, then we can multiply 15 by 194 to find this amount of miles.

15 times 194 equals 2910, so with 15 battery charges, he would go 2910 miles.

#teamtrees #WAP (Water And Plant) #ELM (Every Life Matters)

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15 In²

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1/2 × 6 × 5 = 15

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Step-by-step explanation:

Let the plane through the 4 ft X 4ft square at the top of the figure break the figure into 2 boxes, one has dimensions (8-7) by 3 by 4 = 1ft X 3ft X 4ft, and the other has dimensions 7 by 7 by 4 = 7ft X 7ft X 4ft.

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2 years ago
A submarine on the surface of the ocean descended at a rate of 7 feet per second for 2 minutes. Then it ascended at a rate of 4
kotykmax [81]

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8 0
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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

\\  \\

<h3>To find:</h3>
  • Area of triangle

\\  \\

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} }

\\  \\

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}  \\

\\  \\

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

\\  \\

<h3>know more :-</h3>

\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}]

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make the denominators the same so you cancompare

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