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docker41 [41]
2 years ago
5

Point (-1,3) lies in which quadrant ?​

Mathematics
2 answers:
777dan777 [17]2 years ago
5 0

Quadrant II

Or the top left

Vera_Pavlovna [14]2 years ago
4 0

Answer:

(-1,3) lies in the 2nd quadrant.

Explanation:

Hope you're having a <em>splendiferous</em> day!

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∡C and ∡D are supplementary angles. If m∡D is nine less than twice m∡C, find m∡D.
Gala2k [10]

Answer:

D is 117

Step-by-step explanation:

Let the measure of angle C be x

The measure of D is 9 less than twice C

Mathematically that is 2x-9

If both are supplementary, they add up to be 180

Thus;

x + 2x - 9 = 180

3x = 180 + 9

x = 189/3

x = 63

Recall;

D = 2x-9= 2(63) -9 = 126 -9 = 117

6 0
3 years ago
Jared and Pedro walk 1 mile in about 15 minutes. They can keep up pace for several hours. About how far they walk in 65 minutes
VladimirAG [237]
<h2><u>Answer:</u></h2>

Jared and Pedro walked a total of <u>4.3 miles</u>

<h2><u>Step-by-step explanation:</u></h2>
  • The equation for distance is <u>speed = distance ÷ time.</u>
  • Rearrange the equation to make distance the subject:

        <em>(multiply both sides by time)</em>

<em>         </em><u>Distance = Speed × Time</u>

  • Convert your numbers into suitable units:

        1 mile ÷ 15 = <u>0.06m/minute</u>

  • Finally, substitute the numbers into the equation and solve:

        Distance = 0.06 × 65 = <u>4.3miles</u> (rounded to the nearest decimal place)

<h2 />

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3 years ago
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Find the vertices and foci of the hyperbola with equation quantity x minus 5 squared divided by 144 minus the quantity of y minu
Serjik [45]

\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad  \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a,  k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases}\\\\ -------------------------------


\bf \cfrac{(x-5)^2}{144}-\cfrac{(y-4)^2}{81}=1\implies \cfrac{(x-5)^2}{12^2}-\cfrac{(y-4)^2}{9^2}=1 \\\\\\ \begin{cases} h=5\\ k=4\\ a=12\\ b=9 \end{cases}\implies c=\sqrt{144+81}\implies c=\sqrt{225}\implies c=15 \\\\\\ \stackrel{center}{(5,4)}\qquad \qquad \stackrel{\textit{because is a horizontal traverse hyperbola}}{\stackrel{foci}{\stackrel{(5\pm 15,4)}{(20,4),(-10,4)}}\qquad \qquad \stackrel{vertices}{\stackrel{(5\pm 12,4)}{(17,4),(-7,4)}}}

5 0
3 years ago
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