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TEA [102]
3 years ago
6

What's .60as a fraction in simpolest form

Mathematics
1 answer:
Andreyy893 years ago
4 0
3/5
.60 is equal to 6/10
you can simplify that to 3/5
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Which ratio is not equivalent to the other three?<br> A. 6/15 B 18/45 C 12/25 D 2/5
Sedaia [141]

Step-by-step explanation:

C that is 12/25.

others are × by 3

5 0
3 years ago
Read 2 more answers
The Green family is a family of six people. They have used 4,885.78 gallons of water so far this month. They cannot exceed 9,750
IRINA_888 [86]

Answer:

C. 6x+4885.78<9750.05

=

Step-by-step explanation:

7 0
3 years ago
|0.7x+5|&gt;6.7<br><br>Help plz
Pavlova-9 [17]

Answer:

\large\boxed{x\dfrac{17}{7}}\\\downarrow\\\boxed{x\in\left(-\infty,\ -\dfrac{117}{7}\right)\ \cup\ \left(\dfrac{17}{7};\ \infty\right)}

Step-by-step explanation:

|0.7x+5|>6.7\iff0.7x+5>6.7\ or\ 0.7x+56.7\qquad\text{subtract 5 from both sides}\\0.7x>1.7\qquad\text{divide both sides by 0.7}\\x>\dfrac{1.7}{0.7}\to x>\dfrac{17}{7}\\\\(2)\\0.7x+5

5 0
3 years ago
For this section, using the diagram below, classify the angle pairs as
coldgirl [10]

Answer/Step-by-step explanation:

Angle 1 and angle 11 = none

No special relation between both angles.

Angle 1 and angle 6 = none

Both angles also do not have any special relationship

Angle 12 and angle 15 = same side interior

Angle 12 and angle 15 both interior angles that lie on the same side of the transversal that cuts across two straight lines. They are supplementary.

7 0
4 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
Read 2 more answers
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