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3 years ago

If

Mathematics

1 answer:

andrew-mc [135]3 years ago

8 0

Tan t= 8/15
cot t= 15/8
csc t=-17/8
use sohcahtoa and the reciprocals

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Ayudaaaaaaaaaaaaaaaaaa aaa aaa aaa Doy corona

zvonat [6]

Answer:

The IQR describes the middle 50% of values when ordered from lowest to highest. To find the interquartile range (IQR), first find the median (middle value) of the lower and upper half of the data. These values are quartile 1 (Q1) and quartile 3 (Q3). The IQR is the difference between Q3 and Q1.

Step-by-step explanation:

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3 years ago

1.2 × 10e4− 1.5 × 10e3− 2.2 × 10e2 in scientific notation

madreJ [45]

Answer: 4.5×10^9

Step-by-step explanation:

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the

. The sign of the exponent will depend on the direction you are moving the decimal.

4.5

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3 years ago

Read 2 more answers

Pretzels are given away for free at a local store. The tray is constantly replenished as people take the pretzels.The number of

Tanzania [10]

Answer:

The answer is " $-13.5 \ / minute \\\\$ "

Step-by-step explanation:

The average rate to the changes of pretzels form $t= 1 \ minutes$ to $t=1.5 \ minutes$ will be:

$=\frac{P(1.5)-p(1)}{1.5-1}\\\\=\frac{(20+(1.5)^2- 8 \cos \pi (1.5))-(20+1^2 -8 \cos \pi)}{0.5}\\\\=\frac{22.25-29}{0.5}\\\\=\frac{-675}{50}\\\\=-13.5 \ / minute \\\\$

7 0

3 years ago

A school reserved a banquet hall for the spring dance. In addition to a $100 deposit, each couple must pay $20. If the total cos

AnnZ [28]

Step-by-step explanation:

1.)20x + 100 = 1440

-Let the number of couples be x. So, the total amount paid by all the couple’s are $20x. School paid $100 as security, so the total amount paid is

$ (20x + 100)

2.) 20x = 1440 - 100 ⇒ 20x = 1040

X = 52

3. ) The security money is $100, so the couples paid $1440- 100 = $1040 in total. As the couples paid $ 20 each than- $\frac{1040}{20}$ = 52

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3 years ago

Write an equation in slope-intercept form for the line that passes through (5,0) and is perpendicular to the line described by y

natta225 [31]

For this case we have to;

We have that an equation in slope-intercept form is given by:

$y = mx + b$

Where:

m is the slope

b is the cut point with the y axis

Also, by definition, two lines are perpendicular when the product of their slopes is -1. That is: $m_ {1} * m_ {2} = - 1$

We have the line as data: $y_ {1} = \frac {-5} {2} x + 6$

Then $m_ {1} = \frac {-5} {2}$

We found $m_ {2}$ :

$m_ {1} * m_ {2} = - 1$

$\frac {-5} {2} * m_ {2} = - 1$

$m_ {2} = \frac {-1} {(\frac {-5} {2})}$

$m_ {2} = \frac {(2) (- 1)} {(- 5) (1)}$

$m_ {2} = \frac {2} {5}$

Thus, $y_ {2} = \frac {2} {5} x_ {2} + b_ {2}$

We must find $b_ {2}$ :

We know that $y_ {2}$ passes through the point $(x_ {2}, y_ {2}) = (5,0)$

We substitute the point in the equation of $y_ {2}$ :

$0 = \frac {2} {5} (5) + b_ {2}\\0 = 2 + b_ {2}\\b_ {2} = - 2$

Thus, $y_ {2} = \frac {2} {5} x_ {2} -2$

Then the equation in slope-intercept for the line that passes through (5,0) and is perpendicular to the line described by $y_ {1} = \frac {-5} {2} x_{1} + 6$ is: $y_ {2} = \frac {2} {5} x_ {2} -2$

Answer:

$y_ {2} = \frac {2} {5} x_ {2} -2$

7 0

3 years ago