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3 years ago

Estimate first and then solve using standard algorithm. show your rename the divisor as a whole number. 82.14 divided by 0.6

Mathematics

1 answer:

aleksley [76]3 years ago

8 0

Answer:

136.9

Step-by-step explanation:

82.14=136.9

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PROVE :<br><br>sin^2(-300°).cos^2 (120)+ cos^2(-240 ).sin^2(390)=1/4

leva [86]

Step-by-step explanation:

$\sin {}^{2} ( - 300) \cos {}^{2} (120) + \cos {}^{2} ( - 240) \sin {}^{2} (390) = \frac{1}{4}$

$\sin {}^{2} (60) \cos {}^{2} ( {}^{} 120) + \cos {}^{2} (120) \sin {}^{2} (30)$

$\frac{3}{4} \frac{1}{4} + \frac{1}{4} \frac{1}{4}$

$\frac{3}{16} + \frac{1}{16} = \frac{4}{16} = \frac{1}{4}$

3 0

1 year ago

can someone help me with the circumferences and other stuff on this picture please (zoom out to see the full picture)

Daniel [21]

The corcumference C of a circle is C= 2 x pie x radius
So radius= C/(2pie)

Volume of cylinder= pie x r(squared) x h

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2 years ago

What happens to the graph f(x) = |x/, when f(x) is replaced by 0.3 f(x) +9

Elodia [21]

Answer:

Step-by-step explanation:

First, the graph is flattened, and secondly, the whole resulting graph is translated upward by 9 units.

4 0

3 years ago

Simplify: 20 sin(2x) cos(2x)

alexandr1967 [171]

Use the the double angle formula:
sin(2A)=2sin(A)cos(A)

substitute 2x for A, then
20sin(2x)cos(2x)=10(sin(2(2x))cos(2(2x))=10sin(4x)

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2 years ago

If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w

MatroZZZ [7]

Let $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11}$ be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

$\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.$

2. If the mean weight of the 7 other players is 202 lb, then

$\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.$

3. From the previous statements you have that

$x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.$

Add these two equalities and then divide by 11:

$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.$

Answer: the mean weight of the 11-person team is $208\dfrac{10}{11}\ lb.$

4 0

2 years ago