All categories

4 years ago

Which one doesn’t belong with the other three? Explain your reasoning plz plz answer #3

Mathematics

1 answer:

Ahat [919]4 years ago

7 0

The box provides some examples of how exponents can be used to shorten repeated multiplication. The first, third, and fourth boxes all show how the repeated multiplication of 2, 3, and 5 can all be abbreviated as $2^4, 3^2$ and $5^3$ . The only expression which doesn't belong is $3+3+3+3=3(4)$ , because it's showing how repeated addition can be shortened to multiplication.

You might be interested in

Find the surface area of the figure, with a height of 2 inches.

Vinil7 [7]

I think 128. Surface Area = 2×(2×2 + 2×15 + 2×15) = 128

8 0

3 years ago

You select a marble without looking and then put it back. If you do this 60 times, what is the best prediction possible for the

Damm [24]

Answer:

Step-by-step explanation:

6 0

3 years ago

The length and width of a rectangle are measured as 30cm and 24cm, respectively, with an error in measured of

larisa [96]

Answer:

The maximum error in the calculated area of rectangle is 5.4 cm².

Step-by-step explanation:

Given : The length and width of a rectangle are measured as 30 cm and 24 cm, respectively, with an error in measured of at most 0.1 cm in each.

To find : Use differentials to estimate the maximum error in the calculated area of rectangle ?

Solution :

The area of the rectangle is $A=l\times w$

The derivative of the area is equal to the partial derivative of area w.r.t. length times the change in length plus the partial derivative of area w.r.t. width times the change in width.

i.e. $dA=\frac{\partial A}{\partial L}\Delta L+\frac{\partial A}{\partial W}\Delta W$

Here, $\frac{\partial A}{\partial L}=30,\ \frac{\partial A}{\partial W}=24 ,\ \Delta L=0.1,\ \Delta W=0.1$

Substitute the values,

$dA=(30)(0.1)+(24)(0.1)$

$dA=3+2.4$

$dA=5.4$

Therefore, the maximum error in the calculated area of rectangle is 5.4 cm².

7 0

3 years ago

Academy X + v

bearhunter [10]

A+B=C your lusjis Janis

4 0

3 years ago

Harold uses the binomial theorem to expand the binomial (3x^5 - 1/9y^3)^4

riadik2000 [5.3K]

<h3>The complete question:</h3>

Harold uses the binomial theorem to expand the binomial $(3x^5 -\dfrac{1}{9}y^3)^4$

(a) What is the sum in summation notation that he uses to express the expansion?

(b) Write the simplified terms of the expansion.

Answer:

(a). $(3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}( -\dfrac{1}{9}y^3)^k $$$

(b). $(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561}$

Step-by-step explanation:

(a).

The binomial theorem says

$(x+y)^n=$$\sum_{k=0}^{n} \binom{n}{k}x^{n-k}y^k $$$

For our binomial this gives

$\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n} \binom{4}{k}x^{4-k}y^k $$}$

(b).

We simplify the terms of the expansion and get:

$$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}y^k $$= \binom{4}{0}(3x^5)^{4-0}(-\dfrac{1}{9}y^3 )^0+\binom{4}{1}(3x^5)^{4-1}(-\dfrac{1}{9}y^3 )^1+\\\\\binom{4}{2}(3x^5)^{4-2}(-\dfrac{1}{9}y^3 )^2+\binom{4}{3}(3x^5)^{4-3}(-\dfrac{1}{9}y^3 )^3+\binom{4}{4}(3x^5)^{4-4}(-\dfrac{1}{9}y^3 )^4$

$$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}(-\frac{1}{9}y^3 )^k $$= (3x^5)^{4}+4(3x^5)^{3}(-\frac{1}{9}y^3 )+6(3x^5)^{2}(-\frac{1}{9}y^3 )^2+\\\\4(3x^5)(-\frac{1}{9}y^3 )^3+(-\frac{1}{9}y^3 )^4$

$\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561} }$

3 0

3 years ago