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3 years ago

9

Jane Hilman went to her bank. She had a balance of $1,009.88 in her savings account. She withdrew $130.00 and the teller credite

d her account with $6.19. What is her new balance?

1 answer:

frez [133]3 years ago

3 0

Her new balance is 886.07 dollars hope this helps

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Need answer please show work

Jet001 [13]

Let us set up some variabe:

height: h
base: b

Use the known information:

b = 2h + 8

Now lets find the area

Area = (1/2) * b *h = (1/2) * h * (2h + 8)

Hope that helps!

7 0

2 years ago

PLEASE HELP NEED DONE ASAP !! I WILL REWARD BRAINLIEST

strojnjashka [21]

39,916,800 is the answer

4 0

4 years ago

Mr. Jones took a survey of college students and found that 60 out of 65 students are liberal arts majors. If a college has 8,943

Anit [1.1K]

Should be 5812.95 Students

8 0

4 years ago

matrenka [14]

Answer:

4. Slope of function B = -slope of function A

Step-by-step explanation:

Given:

Function A is given as:

$F(x)=-2x+1$

The above equation is of the form $y=mx+b$ , where $m$ represents slope of the line.

Therefore, on comparing the function A with the above standard form, er conclude that, slope of function A is -2.

Now, from the graph of function, we consider any two points on the graph and determine the slope of the line using the two points.

Let us consider the points $(x_1,y_1)=(1.5,0)\ and\ (x_2,y_2)=(3,3)$

Now, the slope of the line passing through these two points is given as:

$m_B=\frac{y_2-y_1}{x_2-x_1}=\frac{3-0}{3-1.5}=\frac{3}{1.5}=2$

Therefore, slope of function B is 2.

Therefore, the correct relation between the slopes of the two functions is that the slope of function B is negative of the slope of function A.

$m_B=-(m_A)=-(-2)=2$

5 0

3 years ago

[20points][Precal] Find the cross product of -(3/4)V and (1/2)w if v = <-2, 12, -3> and w = <-7, 4, -6>

Lana71 [14]

$\mathbf v=\langle-2,12,-3\rangle=-2\,\vec i+12\,\vec j-3\,\vec k$
$-\dfrac34\mathbf v=\dfrac32\,\vec i-9\,\vec j+\dfrac94\,\vec k$

$\mathbf w=\langle-7,4,-6\rangle=-7\,\vec i+4\,\vec j-6\,\vec k$
$\dfrac12\mathbf w=-\dfrac72\,\vec i+2\,\vec j-3\,\vec k$

$-\dfrac34\mathbf v\times\dfrac12\mathbf w=\begin{vmatrix}\vec i&\vec j&\vec k\\\frac32&-9&\frac94\\-\frac72&2&-3\end{vmatrix}$
$=\left((-9)\times(-3)-\dfrac94\times2\right)\,\vec i-\left(\dfrac32\times(-3)-\dfrac94\times\left(-\dfrac72\right)\right)\,\vec j+\left(\dfrac32\times2-(-9)\times\left(-\dfrac72\right)\right)\,\vec k$
$=\dfrac{45}2\,\vec i-\dfrac{27}8\,\vec j-\dfrac{57}2\,\vec k=\left\langle\dfrac{45}2,-\dfrac{27}8,-\dfrac{57}2\right\rangle$

6 0

3 years ago