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2 years ago

Add/sub the polynomial (1-8x^2+3x)+(1+6x^2+x)

Mathematics

1 answer:

Scorpion4ik [409]2 years ago

3 0

Answer:

-2x² + 4x + 2

Step-by-step explanation:

(1-8x²+3x)+(1+6x²+x)

= 1 - 8x² + 3x + 1 + 6x² + x

= -2x² + 4x + 2 [-2(x² - 2 - 1)]

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There is a quadrilateral MNPQ in which side MN is congruent to side PQ and side NP is parallel to side MQ. The diagonal MP and t

8_murik_8 [283]

Answer: QN = 12

Step-by-step explanation: This quadrilateral is a paralelogram because its 2 opposite sides (NP and MQ) are parallel and the other 2 (MN and PQ) are congruent.

In paralelogram, diagonals bisect each other, which means QR = RN.

If QR = RN:

QR = 6

Then,

QN = QR + RN

QN = 6 + 6

QN = 12

The diagonal QN of quadrilateral MNPQ is QN = 12.

5 0

3 years ago

Please can someone explain to me the steps I have to do to get the answer?

kompoz [17]

Answer:

1/35

Step-by-step explanation:

1/5 divided by 7 is the same as 1/5 x 1/7

=> 1/5 x 1/7 = 1/35

Or using the picture it would be :-

1/5 divided by 7/1

5 0

2 years ago

Read 2 more answers

The graph of the function f(x)=4/5 sqrt x is shown. What is the domain of the function?

S_A_V [24]

Answer:

All real number greater than equal to zero.

Step-by-step explanation:

The function is given by

$f(x) = \frac{4}{5}\sqrt x$

The domain is defined as the input values so that the function is well defined.

here, the values of x should be all real number and zero also.

So, the correct option is (d).

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2 years ago

What is 24 1/4 % expressed as fraction?

VashaNatasha [74]

Answer:

97/4

Step-by-step explanation:

3 0

3 years ago

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Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat

nirvana33 [79]

Answer:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

$\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

And solving for $\frac{dh}{dt}$ we got:

$\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}$

We need to convert the rate given into m^3/min and we got:

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

5 0

3 years ago