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Zigmanuir [339]
3 years ago
7

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
8 0
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)

8 0
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
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