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2 years ago

6

Manny has just filled out his deposit ticket for his savings account. He hands it to the teller and the teller hands it back to

Manny. The teller explains to Manny that the deposit ticket cannot be accepted in its current form. What must Manny do to the deposit ticket in order for the teller to process the deposit ticket?

2 answers:

FinnZ [79.3K]2 years ago

5 0

Answer:

Manny forgot to sign the deposit ticket for less cash back

Step-by-step explanation:

anyanavicka [17]2 years ago

3 0

He has to put how much he is depositing

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Please help me it’s due today

balu736 [363]

The answer should be 15 times 9 equals four

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3 years ago

The endpoints of AB are A(2, 3) and B(8, 1). The perpendicular bisector of AB is CD, and point C lies on AB. The length of CD is

marysya [2.9K]

Comment
If C lies on AB Then it must be the midpoint of AB because CD is the Perpendicular Bisector of AB

Midpoint
A(2,3) B(8,1)
Midpoint formula = (x1 + x2)/2 + (y1 + y2)/2
Midpoint = (2 + 8)/2 + (3 + 1) / 2
Midpoint = 10/2 + 4/2
Midpoint = (5,2) <<<<< answer

Slope AB
slope = (y2 - y1)/(x2 - x1)
slope = (3 - 1) / (2 - 8)
slope = 2/-6 = -1/3

Slope CD
CD_slope * AB_slope = - 1
CD_slope * -1/3 = -1 multiply both sides by - 3
CD_slope * 1 = - 1 * -3
CD_slope = 3 <<<<< Slope CD Answer

Equation CD
Given the midpoint of AB which is C( and the slope of CD
y = 3x + b
2 = 3*5 + b
b = - 13
Equation y = 3x - 13

Using Distance to get D
d^2 = (x1 - x2)^2 + (y2 - y1)'^2
d^2 = (x2 - 5)^2 + (y2 - 2)^2
y2 = 3*x2 - 13
10 = x2^2 - 10x2 + 25 + (3x2 - 13 - 2)^2
10 = x2^2 - 10x2 + 25 + (3x2 - 15)^2
10 = x1^2 - 10x2 + 25 + 9x^2 - 90x2 + 225
10 = 10x2^2 - 100x2 +250
0 = 10x2^2 - 100x2 + 240 Divide through by 10
0 = x2^2 - 10x^2 + 24
0 = (x2 - 6)(x2 - 4)

x2 = 6 or
x2 = 4

Using y2 = 3x - 13
for x2 = 4
then y2 = 3(4) - 13
y2 = - 1

for x2 = 6
y2 = 3*6 - 13
y2 = 5

Possible Answers for D
D = (4,-1)
D = (6,5)

I've included a graph to show a graphical solution.

Please note in a week's time, if no one else answers, you can award a Brainly. This question took over an hour. If someone else answers, please choose one of us.

7 0

3 years ago

Find the exact value of the expression.<br> tan( sin−1 (2/3)− cos−1(1/7))

Sonja [21]

Answer:

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

Step-by-step explanation:

I'm going to use the following identity to help with the difference inside the tangent function there:

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

Let $a=\sin^{-1}(\frac{2}{3})$ .

With some restriction on $a$ this means:

$\sin(a)=\frac{2}{3}$

We need to find $\tan(a)$ .

$\sin^2(a)+\cos^2(a)=1$ is a Pythagorean Identity I will use to find the cosine value and then I will use that the tangent function is the ratio of sine to cosine.

$(\frac{2}{3})^2+\cos^2(a)=1$

$\frac{4}{9}+\cos^2(a)=1$

Subtract 4/9 on both sides:

$\cos^2(a)=\frac{5}{9}$

Take the square root of both sides:

$\cos(a)=\pm \sqrt{\frac{5}{9}}$

$\cos(a)=\pm \frac{\sqrt{5}}{3}$

The cosine value is positive because $a$ is a number between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ because that is the restriction on sine inverse.

So we have $\cos(a)=\frac{\sqrt{5}}{3}$ .

This means that $\tan(a)=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}$ .

Multiplying numerator and denominator by 3 gives us:

$\tan(a)=\frac{2}{\sqrt{5}}$

Rationalizing the denominator by multiplying top and bottom by square root of 5 gives us:

$\tan(a)=\frac{2\sqrt{5}}{5}$

Let's continue on to letting $b=\cos^{-1}(\frac{1}{7})$ .

Let's go ahead and say what the restrictions on $b$ are.

$b$ is a number in between 0 and $\pi$ .

So anyways $b=\cos^{-1}(\frac{1}{7})$ implies $\cos(b)=\frac{1}{7}$ .

Let's use the Pythagorean Identity again I mentioned from before to find the sine value of $b$ .

$\cos^2(b)+\sin^2(b)=1$

$(\frac{1}{7})^2+\sin^2(b)=1$

$\frac{1}{49}+\sin^2(b)=1$

Subtract 1/49 on both sides:

$\sin^2(b)=\frac{48}{49}$

Take the square root of both sides:

$\sin(b)=\pm \sqrt{\frac{48}{49}$

$\sin(b)=\pm \frac{\sqrt{48}}{7}$

$\sin(b)=\pm \frac{\sqrt{16}\sqrt{3}}{7}$

$\sin(b)=\pm \frac{4\sqrt{3}}{7}$

So since $b$ is a number between $0$ and $\pi$ , then sine of this value is positive.

This implies:

$\sin(b)=\frac{4\sqrt{3}}{7}$

So $\tan(b)=\frac{\sin(b)}{\cos(b)}=\frac{\frac{4\sqrt{3}}{7}}{\frac{1}{7}}$ .

Multiplying both top and bottom by 7 gives:

$\frac{4\sqrt{3}}{1}= 4\sqrt{3}$ .

Let's put everything back into the first mentioned identity.

$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$

$\tan(a-b)=\frac{\frac{2\sqrt{5}}{5}-4\sqrt{3}}{1+\frac{2\sqrt{5}}{5}\cdot 4\sqrt{3}}$

Let's clear the mini-fractions by multiply top and bottom by the least common multiple of the denominators of these mini-fractions. That is, we are multiplying top and bottom by 5:

$\tan(a-b)=\frac{2 \sqrt{5}-20\sqrt{3}}{5+2\sqrt{5}\cdot 4\sqrt{3}}$

$\tan(a-b)=\frac{2\sqrt{5}-20\sqrt{3}}{5+8\sqrt{15}}$

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