What is 1.6 powered by 2

Could you help me solve?

Ilya [14]

Answer:

1.14200738982×10^26

Step-by-step explanation:

Substitution can make this integral much easier to evaluate.

<h3>Substitution</h3>

Let u = 7x² -x. Then du = (14x -1)dx. The limits on x become different limits for u:

for x = 1: u = 7(1²) -1 = 6

for x = 3: u = 7(3²) -3 = 60

<h3>Integral</h3>

$\displaystyle\int_1^3{(14x-1)e^{(7x^2-x)}}\,dx=\int_6^{60}{e^u}\,du=\left.e^u\right|^{60}_6\\\\=e^{60}-e^6\approx e^{60}\approx\boxed{1.14200738982\times10^{26}}$

8 0

2 years ago

Find the volume?<br><br> help

olasank [31]

Answer:The answer can be calculated by doing the following steps;

Step-by-step explanation:

5 0

4 years ago

At the begging of the month a store had a balance of -554

makvit [3.9K]

What is the question? Do you mean "At the beginning of the month a store had the balance of $554"? If so, what is the question?

5 0

4 years ago

Factor completely <br> 4(x+1)^2/3 + 12(x+1)^-1/3

wolverine [178]

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
\left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}
\\\\
-------------------------------\\\\
%4(x+1)^2/3 + 12(x+1)^-1/3
4(x+1)^{\frac{2}{3}}+12(x+1)^{-\frac{1}{3}}\implies 4(x+1)^{\frac{2}{3}}+12\cfrac{1}{(x+1)^{\frac{1}{3}}}
\\\\\\$

$\bf 4(x+1)^{\frac{2}{3}}+\cfrac{12}{(x+1)^{\frac{1}{3}}}\impliedby \textit{so, our LCD is }(x+1)^{\frac{1}{3}}
\\\\\\
\cfrac{4(x+1)^{\frac{2}{3}}\cdot (x+1)^{\frac{1}{3}}+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+1)^{\frac{2}{3}+\frac{1}{3}}+12}{(x+1)^{\frac{1}{3}}}
\\\\\\
\cfrac{4(x+1)^{\frac{3}{3}}+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+1)+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4x+4+12}{(x+1)^{\frac{1}{3}}}
\\\\\\
\cfrac{4x+16}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+4)}{\sqrt[3]{x+1}}$

3 0

4 years ago

Read 2 more answers

Please help me with this

postnew [5]

Answer:

(c) BC ≅ BC, reflexive property

Step-by-step explanation:

The conclusion of this proof derives from CPCTC and the SAS congruence postulate. In order for SAS to apply, corresponding sides and the angle between them must be shown to be congruent. The congruence statement ...

ΔABC ≅ ΔDCB

tells you these pairs of sides and angles are congruent:

AB ≅ DC . . . . statement 2
∠ABC ≅ ∠DCB . . . . statement 4
BC ≅ CB . . . . (missing statement 5)
AC ≅ DB . . . . statement 7

That is, the statement needed to complete the proof is a statement that segment BC is congruent to itself. That congruence is a result of the reflexive property of congruence.

4 0

2 years ago