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Nadusha1986 [10]

3 years ago

12

PLEASE HELP (50 Points)....

2 answers:

9966 [12]3 years ago

8 0

otal deposit = 645.30 + 645.30 = $1290.60

Total withdrawals= 158.25 +168.36+157.42+148.46 = $632.49

Balance = 398.26 + 1290.60 - 632.49 = $1056.37

Vedmedyk [2.9K]3 years ago

4 0

Total deposit = 645.30 + 645.30 = $1290.60

Total withdrawals= 158.25 +168.36+157.42+148.46 = $632.49

Balance = 398.26 + 1290.60 - 632.49 = $1056.37

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Which expression is equivalent to (4x^4/2x^3)

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Answer:

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Step-by-step explanation:

Given

$\frac{4x^4}{2x^3}$

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What is 3.1 × 2.57? (6-6)<br> A 8.432<br> B 7.967<br> C 6.621<br> D 5

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Step-by-step explanation:

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2 years ago

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Two vectors A⃗ and B⃗ are at right angles to each other. The magnitude of A⃗ is 5.00. What should be the length of B⃗ so that th

creativ13 [48]

Answer:

Length of B is 7.4833

Step-by-step explanation:

The vector sum of A and B vectors in 2D is

$C=A+B=(a_1+b_1,a_2+b_2)$

And its magnitude is:

$C=\sqrt{(a_1+b_1)^2+(a_2+b_2)^2} =9$

Where

$a_1=Asinx$

$a_2=Acosx$

$b_1=Bsin(x+90)$

$b_2=Bcos(x+90)$

Using the properties of the sum of two angles in the sin and cosine:

$b_1=Bsin(x+90)=B(sinx*cos90+sin90*cosx)=Bcosx$

$b_2=Bcos(x+90)=B(cosx*cos90-sinx*sin90)=-Bsinx$

Sustituying in the magnitud of the sum

$C=\sqrt{(Asinx+Bcosx)^2+(Acosx-Bsinx)^2} =9$

$C=\sqrt{A^2sin^2x+2ABsinxcosx+B^2cos^2x+A^2cos^2x-2ABsinxcosx+B^2sin^2x} =9$

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$C=\sqrt{A^2+B^2} =9$

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$A^2+B^2 =9^2$

$B^2 =9^2-A^2$

Sustituying the value of the magnitud of A

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