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2 years ago

What value for A makes this equation true? 4 • 48 = (4 • 40) (4 • B)

Mathematics

1 answer:

Katena32 [7]2 years ago

4 0

$4\cdot48=(4\cdot40)(4\cdot B)\\
192=160\cdot4B\\
640B=192\\
B=\dfrac{192}{640}=\dfrac{3}{10}

$

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What type of covalent bond diagram only shows the outer shell of electrons?

vitfil [10]

Correct answer: Dot-and-cross-diagram

Dot-and-cross diagrams are used to represent covalent bonds. The shared electron from one atom is shown as a dot, while the shared electron from the other atom is shown as a cross.

When drawing dot-and-cross diagrams for covalent bonds, you only need to show the electrons in the highest occupied energy level, as only these are involved.

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3 years ago

nevsk [136]

#Personal computer:-

Total=480

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$\\ \bull\sf\dashrightarrow \dfrac{480}{1500}\times 100$

$\\ \bull\sf\dashrightarrow 32\%$

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$\\ \bull\sf\dashrightarrow 0.26(1500)$

$\\ \bull\sf\dashrightarrow 390$

#Wireless communication:-

$\\ \bull\sf\dashrightarrow 0.18(1500)$

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#Television:-

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$\\ \bull\sf\dashrightarrow \dfrac{150}{1500}\times 100$

$\\ \bull\sf\dashrightarrow 10\%$

#Others

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2 years ago

Read 2 more answers

What is the answer? 1/4 7/20 5/11 1/2

julsineya [31]

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2 years ago

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Factor the following: 1. 5(a+3)²- (a+3 2. p²- 6 - q · (p²- 6)²

scoundrel [369]

Answer:

1. $(a + 3)(5a-14)$

2. $(p^2-6)[1-q(p^2-6)]$

Step-by-step explanation:

1. The first thing to do to factor the expression is to take the expression (a + 3) as a common factor with its lowest exponent.

Then the expression. $5(a + 3)^2- (a + 3)$ remains as:

$(a + 3)(5 (a + 3) -1)\\\\(a + 3)(5a +15 -1)\\\\(a + 3)(5a-14)$

2. The first thing to do to factor the expression is to take the expression $p^2 -6$ as its common factor with its lowest exponent.

Then the expression $p^2- 6 - q(p^2- 6)^2$ remains as:

$(p^2-6)[1-q (p^2-6)]$

8 0

2 years ago

how to integrate <img src="https://tex.z-dn.net/?f=e%5E%7B2s%7D%20%2ACos%20%5Cfrac%7Bs%7D%7B4%7D" id="TexFormula1" title="e^{2s}

icang [17]

Answer:

$\int\limits {e^{2s} cos\frac{s}{4} ds$ = $\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s + sin \frac{1}{4} s} ))$

Step-by-step explanation:

Step(i):-

Given that $f(s) = e^{2s} cos\frac{s}{4}$

Now integrating

$\int\limits {f(s)} \, ds = \int\limits {e^{2s} cos\frac{s}{4} ds$

By using integration formula

$\int\limits { e^{ax} cos b x dx = \frac{e^{ax} }{a^{2}+b^{2} } ( a cos b x + b sin b x )$

Step(ii):-

$\int\limits {e^{2s} cos\frac{s}{4} ds$ = $\frac{e^{2s} }{(2)^{2}+(\frac{1}{4}) ^{2} } ( 2 cos (\frac{1}{4} ) s + \frac{1}{4} sin \frac{1}{4} s ))$

= $\frac{e^{2s} }{(4+\frac{1}{16})} ( 2 cos (\frac{1}{4} ) s + \frac{1}{4} sin \frac{1}{4} s ))$

= $\frac{e^{2s} }{(\frac{65}{16} } ( \frac{8 cos (\frac{1}{4} ) s + sin \frac{1}{4} s}{4} ))$

= $16 X\frac{e^{2s} }{65 } ( \frac{8 cos (\frac{1}{4} ) s + sin \frac{1}{4} s}{4} ))$

= $\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s + sin \frac{1}{4} s} ))$

Final answer:-

$\int\limits {e^{2s} cos\frac{s}{4} ds$ = $\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s + sin \frac{1}{4} s} ))$

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2 years ago