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3 years ago

Is -3y = 5x + 4 perpendicular

Mathematics

2 answers:

aivan3 [116]3 years ago

5 0

Answer:

no it is not perpedicular

Step-by-step explanation:

I plugged the equation into a graphing calculator and the line does not appear to be perpendicular

Dahasolnce [82]3 years ago

4 0

Answer:

3y = 5x+4

3y=9x

yx=9/3

yx=3

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Resistors for electronic circuits are manufactured on a high-speed automated machine. The machine is set up to produce a large r

Anettt [7]

Answer:

See explanation

Step-by-step explanation:

Given

See attachment for proper presentation of question

Required

Mean and Range

To do this, we simply calculate the mean and the range of each row.

$\bar x = \frac{\sum x}{n}$ ---- mean

Where:

$n = 4$ ---- number of rows

$R = Highest - Lowest$ --- range

So, we have:

Sample 1

$\bar x_1 = \frac{1027+ 994 +977 +994 }{4}$

$\bar x_1 = 998$

$R_1 = 1027- 994$

$R_1 = 33$

Sample 2

$\bar x_2 = \frac{975 +1013 +999 +1017}{4}$

$\bar x_2 = 1001$

$R_2 = 1017 - 975$

$R_2 = 42$

Sample 3

$\bar x_3 = \frac{988 +1016 +974 +997}{4}$

$\bar x_3 = 993.75$

$R_3 = 1016-974$

$R_3 = 42$

Sample 4

$\bar x_4 = \frac{998 +1024 +1006 +1010}{4}$

$\bar x_4 = 1009.5$

$R_4 = 1024 -998$

$R_4 = 26$

Sample 5

$\bar x_5 = \frac{990 +1012 +990 +1000}{4}$

$\bar x_5 = 998$

$R_5 = 1012 -990$

$R_5 = 22$

Sample 6

$\bar x_6= \frac{1016 + 998 +1001 +1030}{4}$

$\bar x_6= 1011.25$

$R_6= 1030-998$

$R_6= 32$

Sample 7

$\bar x_7 = \frac{1000 +983 +979 +971}{4}$

$\bar x_7 = 983.25$

$R_7 = 1000-971$

$R_7 = 29$

Sample 8

$\bar x_8 = \frac{973 +982 +975 +1030}{4}$

$\bar x_8 = 990$

$R_8 = 1030-973$

$R_8 = 57$

Sample 9

$\bar x_9 = \frac{992 +1028 +991 +998}{4}$

$\bar x_9 = 1002.25$

$R_9 = 1028 -991$

$R_9 = 37$

Sample 10

$\bar x_{10} = \frac{997 +1026 +972 +1021}{4}$

$\bar x_{10} = 1004$

$R_{10} = 1026 -972$

$R_{10} = 54$

Sample 11

$\bar x_{11} = \frac{990 +1021 +1028 +992}{4}$

$\bar x_{11} = 1007.75$

$R_{11} = 1028 -990$

$R_{11} = 38$

Sample 12

$\bar x_{12} = \frac{1021 +998 +996 +970}{4}$

$\bar x_{12} = 996.25$

$R_{12} = 1021 -970$

$R_{12} = 51$

Sample 13

$\bar x_{13} = \frac{1027 +993 +996 +996}{4}$

$\bar x_{13} = 1003$

$R_{13} =1027 -993$

$R_{13} =34$

Sample 14

$\bar x_{14} = \frac{1022 +981 +1014 +983}{4}$

$\bar x_{14} = 1000$

$R_{14} = 1022 -981$

$R_{14} = 41$

Sample 15

$\bar x_{15} = \frac{977 +993 +986 +983}{4}$

$\bar x_{15} = 984.75$

$R_{15} = 993-977$

$R_{15} = 16$

8 0

3 years ago

Marco needs to buy some dog food. At the nearest store, 3 bags of dog food cost $9.75. How much Marco spend on 5 bags of dog foo

zhenek [66]

Answer:

$16.25

Step-by-step explanation:

9.75/3 = 3.25

So therefore 1 bag costs $3.25.

$3.25 × 5 = $16.25

3 0

3 years ago

The quantity, Q, of a drug in the blood stream begins with 250 mg and decays to one-fifth its value over every 90 minute period.

Mazyrski [523]

Answer:

$a=250 \, mg\\\\b=5\\\\T= 90'$

Step-by-step explanation:

We have that $Q(t) = a\cdot b^{-\frac{t}{T}} \\$

Where t is the time (in minutes) and for the sake of dimensional consistency, let's assume that T is also in minutes, b is an adimensional number, and a is in mg. So we will have Q in mg as a consequence.

We now want to find out what values these constants might take. Let's see what happens when $t=0$ , that is, just as we start. At that point, we have that the amount of drug in the bloodstream must be equal to 250mg, thus:

$Q(0)= a\cdot b ^{-\frac{0}{T} }=250\,mg\\Q(0)= a=250\,mg$

We have found the constant $a$ ! It is the initial amount of drug! we have made use of the fact that any number raised to the 0th power is equal to one.

Now, we know that every 90 minutes, the amount of drug decreases to one fifth of its former value. How do we put this in mathematical form? Like so:

$Q(t+90')=Q(t)/5$

That is, 90 minutes after time t the amount of drug will be one fifth of the amount of drug at time t. Let's expand the last equation:

$Q(t+90')=Q(t)/5\\\\a\cdot b^{-\frac{t+90'}{T} }=a\cdot b^{-\frac{t}{T} }/5\\\\ b^{-\frac{t+90'}{T} }=b^{-\frac{t}{T} }/5\\\\b^{-\frac{t}{T} }b^{-\frac{90'}{T} }=b^{-\frac{t}{T} }/5\\\\b^{-\frac{90'}{T} }=\frac{1}{5}$

Now the last expression isn't enough to determine both T and b, but that also means that we have some freedom in how we choose them. What seems most simple is to pick $T=90'$ and thus we will get:

$b^{-1 }=\frac{1}{5}\\\\\frac{1}{b}= \frac{1}{5}\\\\b=5$

And that is our final result.

3 0

3 years ago

Suppose the real risk-free rate is 2.50% and the future rate of inflation is expected to be constant at 7.00%. What rate of retu

AfilCa [17]

Answer:

option (a) 9.50%

Step-by-step explanation:

Data provided in the question:

The real risk free rate = 2.50%

The future rate of inflation = 7.00%

Now,

The Expected rate of return after 5 - year treasury deposited will be calculated using the relation,

Return rate = Inflation rate + Real risk free rate

on substituting the respective values, we get

Return rate = 7.00% + 2.50%

Return rate = 9.50%

Hence,.

the correct answer is option (a) 9.50%

4 0

4 years ago

1. In the following statement, express the first number as a percentage of the second number.

AleksandrR [38]

9514 1404 393

Answer:

120.8%

Step-by-step explanation:

Maybe you want to find $42,900 as a percentage of $35,500.

That is ...

42900/35500 × 100% = 120.8%

<em>Comment on the question</em>

The question is easier to read if you don't shuffle the phrases. Copying and pasting often requires a check and edit step afterward.

3 0

3 years ago