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kolbaska11 [484]
3 years ago
9

John made a lump sum deposit of $ 6,300 in an account that pays 6.5% per year. Find

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
4 0

Answer:  $8,347.50 is the value of his account after five years.

Interest is $409.50 per year.

Total amount of interest paid is $2,047.50.

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John held a garage sale. He priced all the items at a dime or a quarter. His sales totaled
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John held a garage sale. He priced all the items at a dime or a quarter, his sales totaled less than $5, the solution is shaded region in quadrant I.

<h3>What is a slope?</h3>

In mathematics, line's slope, also known as its gradient, is a numerical representation of the line's steepness and direction. The letter m is frequently used to represent the slope; reason for this usage is unclear, but it can be found in O'Brien's (1844) and Todhunter's (1888) formulations of equation for a straight line as "y = mx + b" and "y = mx + c," respectively.

He priced all items at a dime or a quarter. His sales totaled less than $5

Since dime =$ 0.10=$0.10 and the quarter =$0.25=$0.25

Let x=x= the number of the items sold for a dime

Let y=y= the number of the items sold for a quarter

0.10x+0.25y<5

0.10x+0.25y<5

Inequality in the slope-intercept form

0.10x+0.25y Subtract 0.10x from both sides

0.25y<-0.10x+5 Divide both sides by 0.25

Subtract 0.10x from both sides

y= -\frac{2}{5}x+20\\

The slope is, -\frac{2}{5} and the y- intercept is 20

Since inequality sign is << so, draw a dashed line

Test point (0,0)(0,0)

Substituting the values 0 for x and 0 for y,

y < -\frac{2}{5}x+20\\\\0 < -\frac{2}{5}(0)+20\\\\0 < +20

Since inequality is true

So, (0,0)(0,0) is solution of the inequality

Shade region containing (0,0)(0,0).

The graph below shows all possible solutions

Since boundary line is dashed and the number of items is always positive

The solution is shaded region in quadrant I

To know more about quadrant visit: brainly.com/question/7196312

#SPJ9

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1 year ago
20 POINTS!!!
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\dfrac{2}{\sqrt3\cos x+\sin x}=\sec\left(\dfrac{\pi}{6}-x\right)\\\\\dfrac{2}{\sqrt3\cos x+\sin x}=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}\ \ \ \ \ (*)\\----------------\\\\\cos\left(\dfrac{\pi}{6}-x\right)\ \ \ \ |\text{use}\ \cos(x-y)=\cos x\cos y+\sin x\sin y\\\\=\cos\dfrac{\pi}{6}\cos x+\sin\dfrac{\pi}{6}\sin x=\dfrac{\sqrt3}{2}\cos x+\dfrac{1}{2}\sin x=\dfrac{\sqrt3\cos x+\sin x}{2}\\------------------------------

(*)\\R_s=\sec\left(\dfrac{\pi}{6}-x\right)=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}=\dfrac{1}{\dfrac{\sqrt3\cos x+\sin x}{2}}\\\\=\dfrac{2}{\sqrt3\cos x+\sin x}=L_s

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:pog_champ:

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