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stepladder [879]
3 years ago
8

If a bicyclist travels 3 miles in 15 minutes, how far will he travel in 25 minutes maintaining a constant speed? (includes units

)
Mathematics
2 answers:
marysya [2.9K]3 years ago
5 0
15min / 3miles = 1mile per 5 min

So 25min / 5min = 5miles

Answer is 5 miles in 25 minutes
Irina-Kira [14]3 years ago
3 0

<u>Answer:</u>

5 miles

<u>Step-by-step explanation:</u>

15 × ? = 25

3 × ? = ?

15 × 1 and 2/3 = 25

3 × 1 and 2/3 = ?

3 × 1 and 2/3 = 5

Your welcome and Rate as Brainliest

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Which ordered pair does NOT lie on the graph of h(x)=-2x + 7
Rus_ich [418]

Answer:

A

Step-by-step explanation:

To determine if the ordered pairs lie on the graph, substitute the x- coordinate of the point into the equation and if the value agrees with the y- coordinate of the point then it lies on the graph

A

x = - 2 : y = -2(- 2) + 7 = 4 + 7 = 11 ≠ 3 ← (-2,3) not on graph

B

x = - 1 : y = - 2(- 1) + 7 = 2 + 7 = 9 ← (- 1, 9) lies on graph

C

x = 3 : y = -2(3) + 7 = - 6 + 7 = 1 ← (3, 1) lies on graph

D

x = 4 : y = - 2(4) + 7 = - 8 + 7 = - 1 ← (4, - 1) lies on graph

8 0
2 years ago
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243.875 to ne rest tenth, hundredth, ten and hundred
shutvik [7]
243.9 tenth
243.88 for hundredth
240 for ten
200 for hundred
6 0
3 years ago
Find the ratio of the perimeter of the first figure to the perimeter of the second.
ololo11 [35]

Answer:

2:1

Step-by-step explanation:

The perimeter of the first is twice as large as the 2nd

5 0
2 years ago
Passing through (-2,1 ) and perpendicular to<br> 4x + 7y + 3 = 0.
Lunna [17]

Answer:

<h2>7x - 4y + 18 = 0</h2>

Step-by-step explanation:

The slope-intercept form of an equation of a line:

y=mx+b

m - slope

b - y-intercept

========================================

Let

k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2

========================================

We have the equation of a line in a general form (Ax + By + C = 0)

Convert it to the slope-intercept form:

4x+7y+3=0             <em>subtract 7y from both sides</em>

4x+3=-7y         <em>divide both sides by (-7)</em>

-\dfrac{4}{7}x-\dfrac{3}{7}=y\to m_1=-\dfrac{4}{7}

Therefore

m_2=-\dfrac{1}{-\frac{4}{7}}=\dfrac{7}{4}

We have the equation:

y=\dfrac{7}{4}x+b

Put the coordinates of the point (-2, 1) to the equation, and solve for <em>b</em> :

1=\dfrac{7}{4}(-2)+b

1=-\dfrac{7}{2}+b     <em>multiply both sides by 2</em>

2=-7+2b           <em>add 7 to both sides</em>

9=2b            <em>divide both sides by 2</em>

[te]x\dfrac{9}{2}=b\to b=\dfrac{9}{2}[/tex]

Finally:

y=\dfrac{7}{4}x+\dfrac{9}{2} - <em>slope-intercept form</em>

Convert to the general form:

y=\dfrac{7}{4}x+\dfrac{9}{2}         <em>multiply both sides by 4</em>

4y=7x+18      <em>subtract 4y from both sides</em>

0=7x-4y+18

6 0
2 years ago
Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess
Zinaida [17]

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill

\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2

4 0
3 years ago
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