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3 years ago

Find the value of the expression 4a^4 − 2b^2 + 40 when a = 2 and b = 7

Mathematics

2 answers:

Effectus [21]3 years ago

7 0

Answer:6

Step-by-step explanation:

4(2)^4 − 2(7)^2 + 40

4(16)-2(7)^2+40

64-2(49)+40

64-98+40

-34+40

Flura [38]3 years ago

4 0

Answer:

Step-by-step explanation:

4a^4 - 2b^2 + 40

Put a = 2 and b = 7

4(2)^4 - 2(7)^2 + 40

Solve the powers first.

4(16) - 2(49) + 40

Multiply.

64 - 98 + 40

Add and subtract.

= 6

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3 years ago

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Answer:

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3 years ago

A study was conducted on the relationship between speeds of cars and gas mileage. The correlation coefficient was 0.45. Later, t

True [87]

Answer:

0.45

Step-by-step explanation:

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3 years ago

A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$