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mart [117]
3 years ago
14

eric is 70 inches tall pablo is as half as tall as eric annette is 5.25 inches taller than pablo how tall is annette

Mathematics
1 answer:
Kazeer [188]3 years ago
7 0

Answer:

40.25 inches

Step-by-step explanation:

Following what the problem mentions:

"eric is 70 inches tall"

Eric=70in

"pablo is as half as tall as eric"

pablo=\frac{Eric}{2}

pablo=\frac{70in}{2}=35in

and "annette is 5.25 inches taller than pablo"

annete=pablo+5.25in

annette=35in+5.25in =40.45in

Annette is 40.25 inches tall

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What is the geometric mean of 4 and 10
GalinKa [24]

Answer: 6.3

Step-by-step explanation:

The formula for geometric mean is given as :

G.m = \sqrt{pq}

Therefore , the geometric mean between 4 and 10 will be :

G.m = \sqrt{4(10)}

      = \sqrt{40}

      = 2\sqrt{10}

     = 6.3

8 0
3 years ago
The choir teacher is arranging the Junior and Senior choir members in equal rows for an upcoming concert. Each row will contain
ANTONII [103]

The answer is 9

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In step one do you found the volume (in cubic feet) of the main tank(359,006.67). The maximum density of killer whales per cubic
Luba_88 [7]

Answer:

  4 killer whales

Step-by-step explanation:

The dimensional analysis is ...

  (whales/ft³)(ft³/tank) = whales/tank

Putting the numbers with the units, we get ...

  (1.1142·10^-5 whales/ft³)(3.5900667·10^5 ft³/tank) = 4.00005... whales/tank

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6 0
3 years ago
Someone please help me with this lol… have no idea what I’m doing
Sholpan [36]

Given:

\cos \theta =\dfrac{3}{5}

\sin \theta

To find:

The quadrant of the terminal side of \theta and find the value of \sin\theta.

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

\cos \theta =\dfrac{3}{5}

\sin \theta

Here cos is positive and sine is negative. So, \theta must be lies in Quadrant IV.

We know that,

\sin^2\theta +\cos^2\theta =1

\sin^2\theta=1-\cos^2\theta

\sin \theta=\pm \sqrt{1-\cos^2\theta}

It is only negative because \theta lies in Quadrant IV. So,

\sin \theta=-\sqrt{1-\cos^2\theta}

After substituting \cos \theta =\dfrac{3}{5}, we get

\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}

\sin \theta=-\sqrt{1-\dfrac{9}{25}}

\sin \theta=-\sqrt{\dfrac{25-9}{25}}

\sin \theta=-\sqrt{\dfrac{16}{25}}

\sin \theta=-\dfrac{4}{5}

Therefore, the correct option is B.

6 0
2 years ago
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