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AnnZ [28]
2 years ago
6

Add 1/10 + 2/15 (show steps pls-) :)

Mathematics
1 answer:
Ksivusya [100]2 years ago
3 0

Answer:

\frac{7}{30}

OR...

0.233333333

Step-by-step explanation:

Least common multiple of 10 and 15 is 30. Convert \frac{1}{10} and \frac{2}{15}  to fractions with denominator 30.

\frac{3}{30} +\frac{4}{30}

Since \frac{3}{30} and \frac{4}{30} have the same denominator, add them by adding their numerators.

\frac{3+4}{30}

Add 3 and 4 to get 7.

\frac{7}{30}

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Anarel [89]

Answer:

The answer is 8ft

Step-by-step explanation:

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3 years ago
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A cone has a height of 7 ft and a radius of 4 ft. Which equation can find the volume of the cone?
KiRa [710]

Answer:

V = one-third pi (4 squared) (7) feet cubed

Step-by-step explanation:

Volume of cone v = 1/3×π(r²h)

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V = 1/3 × π(4²×7)

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3 years ago
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Which equation represents a hyperbola with a center at (0, 0), a vertex at (−48, 0), and a focus at (50, 0)?
Lerok [7]

bearing in mind that "a" is the length of the traverse axis, and "c" is the distance from the center to either foci.

we know the center is at (0,0), we know there's a vertex at (-48,0), from the origin to -48, that's 48 units flat, meaning, the hyperbola is a horizontal one running over the x-axis whose a = 48.

we also know there's a focus point at (50,0), that's 50 units from the center, namely c = 50.


\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}\\ \textit{asymptotes}\quad y= k\pm \cfrac{b}{a}(x- h) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}


\bf \begin{cases} h=0\\ k=0\\ a=48\\ c=50 \end{cases}\implies \cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{b^2}=1 \\\\\\ c=\sqrt{a^2+b^2}\implies \sqrt{c^2-a^2}=b\implies \sqrt{50^2-48^2}=b \\\\\\ \sqrt{196}=b\implies 14=b~\hspace{3.5em}\cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{14^2}=1\implies \cfrac{x^2}{48^2}-\cfrac{y^2}{14^2}=1

8 0
3 years ago
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Help me and i’ll mark you brainliest plus you get 20 points !!!!
Anton [14]

Answer:

Angle bisector

Step-by-step explanation:

Find the measure of the angle COA. By angle addition postulate,

m\angle COX=m\angle AOX+m\angle COA

From the diagram,

m\angle COX=80^{\circ}\\ \\m\angle AOX=40^{\circ},

then

80^{\circ}=m\angle COA+40^{\circ}\\ \\m\angle COA=80^{\circ}-40^{\circ}=40^{\circ}

Find the measure of the angle BOA. By angle addition postulate,

m\angle BOX=m\angle AOX+m\angle BOA

From the diagram,

m\angle BOX=60^{\circ}\\ \\m\angle AOX=40^{\circ},

then

60^{\circ}=m\angle BOA+40^{\circ}\\ \\m\angle BOA=60^{\circ}-40^{\circ}=20^{\circ}

Find the measure of the angle COB. By angle addition postulate,

m\angle COX=m\angle BOX+m\angle COB

From the diagram,

m\angle BOX=60^{\circ}\\ \\m\angle COX=80^{\circ},

then

80^{\circ}=m\angle COB+60^{\circ}\\ \\m\angle COB=80^{\circ}-60^{\circ}=20^{\circ}

This means, the measures of angles COB and BOA are the same and are equal half the measure of angle COA, so angles COB and BOA are congruent. This means, the ray OB is the angle bisector of angle COA

6 0
3 years ago
A) 240 cm 2<br> B) 270 cm 2<br> C) 318 cm 2 <br> D) 348 cm 2
fgiga [73]
I think the answer is D.348
7 0
4 years ago
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