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3 years ago

6

Help I’m struggling so much in math

1 answer:

andreev551 [17]3 years ago

7 0

Slope = rise/run
rise/run = -8/5

FINAL ANSWER -8/5

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o-na [289]

We know that:
Mean = 82 mm and SD = 10 mm ( standard deviation )
82 - 3 * SD = 82 - 3 * 10 = 82 - 30 = 52 mm
82 + 3 * SD = 82 + 3 * 10 = 82 + 30 = 112 mm
Population between 52 and 112 mm is within +/- 3 standard deviations from the mean.
By the 66- 95 - 99.7 % rule it is: 99.7% of the test group.
0.977 * 500 = 498.5
Answer:
99.7 % of the test group have a diastolic pressure between 52 and 112 mm, or 498 men.

5 0

3 years ago

pioneers are buying mules to trek across the country. each of the 5 wagons require e mules to pull them. write the equation to e

tekilochka [14]

Answer:

d = 5e

Step-by-step explanation:

Number of wagons = 5

Number of miles per wagon = e

Total number of mules = d

Hence, relationship between total mules d needed will be ;

Total mules = number of wagons * Numbe rof mules per wagon

d = 5e

4 0

3 years ago

When a sunfloweris 2 weeks old it is 38 centimeters tall when it is 6 weeks old it is 114 cm tall how tall is it when it is 3 ½

Anna11 [10]

2 wk --- 38 cm
3.5 wk --- x cm
x = 3.5 wk x 38 cm
---------------------
2 wk
x= 66.5 cm

5 0

3 years ago

Write the conversion factor for seconds to minutes. Use the factor to convert 135 seconds to minutes.

OverLord2011 [107]

Answer

Find out the conversion factor for seconds to minutes and convert 135 seconds to minutes.

To prove

1 minute = 60 second

for seconds to minutes.

$1 second = \frac{1}{60}\ minutes$

Therefore the conversion factor for seconds to minutes be

$= \frac{1}{60}\ minutes$

Now convert 135 seconds to minutes.

$= \frac{135\times 1}{60}$

= 2.25 minutes

Hence proved

4 0

3 years ago

Read 2 more answers

Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).

fomenos

Answer:

$\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1$

Step-by-step explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is $2a=|4--10|$ .

$\implies 2a=|14|$

$\implies 2a=14$

$\implies a=7$

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is $(\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)$

We need to use the relation $a^2+b^2=c^2$ to find $b^2$ .

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

$c=|6--3|=9$

$\implies 7^2+b^2=9^2$

$\implies b^2=9^2-7^2$

$\implies b^2=81-49$

$\implies b^2=32$

We substitute these values into the standard equation of the hyperbola to obtain:

$\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1$

$\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1$

7 0

3 years ago