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3 years ago

2/5(a+2b)^2 (if a=3 and b=-4)

Mathematics

1 answer:

kondor19780726 [428]3 years ago

4 0

The answer is 10. Hope this helps.

Multiply the numbers

2/5 •(3-8)^2

Subtract (3-8)

2/5• (-5)^2

A negative base raised to an even power equals a positive

2/5• (5)^2

Reduce the expression with 5.

2•5

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What is the value of A when we rewrite 2^x-6 + 2x as A * 2^x?

Leto [7]

Algebraic expressions are expressions that use numbers and variables

The value of A is 65/64

<h3>How to determine the value of A</h3>

The expression is given as:

$2^{x - 6} + 2^x$

Rewrite the expression as:

$2^{x - 6} + 2^x = \frac{2^x}{2^6} + 2^x$

Factor out 2^x

$2^{x - 6} + 2^x = 2^x(\frac{1}{2^6} + 1)$

Rewrite as product

$2^{x - 6} + 2^x = 2^x * (\frac{1}{2^6} + 1)$

The expression to compare with is given as: A * 2^x.

So, we have:

$A * 2^x =2^x * (\frac{1}{2^6} + 1)$

Divide both sides by 2^x

$A =\frac{1}{2^6} + 1$

Take LCM

$A =\frac{1 + 2^6}{2^6}$

$A =\frac{65}{64}$

Hence, the value of A is 65/64

Read more about expressions at:

brainly.com/question/4344214

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1 year ago

The graph of y=−3x+2

Sophie [7]

The attached image is the graph.

3 0

2 years ago

Read 2 more answers

Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3

Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

$\frac{dV}{dt} = -kA$

Using Euler's method

$\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\$

Where initial droplet volume is:

$V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3$

Hence, the iterative solution will be as next:

i = 1, ti = 0, Vi = 65.45

$V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88$

i = 2, ti = 0.5, Vi = 63.88

$V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33$

i = 3, ti = 1, Vi = 62.33

$V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813$

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

$r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\$

The average change of droplet radius with time is:

Δr/Δt = $\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min$

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

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3 years ago

Which table shows a linear function?<br> Will mark brainliest

IRISSAK [1]

Answer:

All of them

Step-by-step explanation:

none of the graphs repeat any numbers on the x

8 0

3 years ago

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Which values of A and B make the following equations true?

Korolek [52]

<span>(5x^7 y^2)(-4x^4 y^5) =-20x^a y^b
</span><span>
solve this:
(5x^7 y^2)(-4x^4 y^5)
= -20 x^(7 + 4) y^(2 + 5)
= -20 x^11 y^7

answer

</span>A) a=11, b=7

5 0

3 years ago