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2 years ago

Evaluate S(3)=10+20log4(3+1)

1 answer:

4 0

Asssuming you mean

evaluate $10+20log_4(3+1)$

remember $log_a(a)=1$
so
$10+20log_4(3+1)$ =
$10+20log_4(4)$ =
$10+20(1)$ =
10+20=
30

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Which sum will be neither positive nor negative. a. 3+2 b. -4+(-4) c. -6+6 d. 7+(-6)

anastassius [24]

Answer:

Step-by-step explanation:

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2 years ago

What fraction is .675

Vitek1552 [10]

675/1000. Simplified it would be 27/40.

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2 years ago

If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Harlamova29_29 [7]

Answer:

Given: In parallelogram ABCD, AC=BD

To prove : Parallelogram ABCD is rectangle.

Proof : in △ACB and △BDA

AC=BD ∣ Given

AB=BA ∣ Common

BC=AD ∣ Opposite sides of the parallelogram ABCD

△ACB ≅△BDA∣SSS Rule

∴∠ABC=∠BAD...(1) CPCT

Again AD ∥ ∣ Opposite sides of parallelogram ABCD

AD ∥BC and the traversal AB intersects them.

∴∠BAD+∠ABC=180∘ ...(2) _ Sum of consecutive interior angles on the same side of the transversal is

180∘

From (1) and (2) ,

∠BAD=∠ABC=90∘

∴∠A=90∘ and ∠C=90∘

Parallelogram ABCD is a rectangle.

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2 years ago

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$

And,

$f'(A)=\frac{dB}{dA}=\sqrt{3}$ from part b

Therefore, the linear approximation at $A=\frac{\pi}{4}$ is,

$f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}$

Use part (c), when $A=46°$ , B is approximately,

$B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°$

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2 years ago

Students are given 3 minutes to complete each multiple-choice question on a test and 8 minutes for each free-response question.

Alja [10]

The value that could replace variable x in the table is (15 - m)8

<h3 /><h3>How to the value of x in an expression ?</h3>

The students are given 3 minutes to complete each multiple choice question on a test and 8 minutes for each free response question.

From the table, the time to answer the whole multiple choice question is as follows:

total time = 3m

From the table, the total time taken to answer the free response question is as follows:

total time = (15 - m)8

learn more on variable here: brainly.com/question/12798965

#SPJ1

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2 years ago