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dedylja [7]
3 years ago
6

Is it True or false ?

Mathematics
1 answer:
enyata [817]3 years ago
5 0

Answer:True

Perpendicular lines are lines that intersect at a right angle.

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Which is the equation of a line that has a slope of Negative two-thirds and passes through point (–3, –1)?
weqwewe [10]

Answer:

Y=2/3x+1

Step-by-step explanation:

5 0
2 years ago
What are the x and y intercepts of the below equation
lozanna [386]
Plug in 0 for x
0 - 6y = 30
y = -5
Y intercept: (0,-5)

Plug in 0 for y
5x = 30
X = 6
X intercept: (6,0)
3 0
2 years ago
Please help me I don’t understand :(
horrorfan [7]

b=5, commutative property

c=3, associative property

8 0
2 years ago
Cesar is creating a schedule to practice his instrument for band. He needs to practice for a total of 10 hours in a week. On Mon
ratelena [41]

Answer: See explanation

Step-by-step explanation:

Your question isn't really clear. Let me help you rephrase it and solve.

Let's say Cesar is creating a schedule to practice his instrument for band. He needs to practice for a total of 10 hours in a week. On Monday he practice for 1 1/2 hours, on Tuesday he practiced for 2 1/4 hours, and on Wednesday he practiced for 1 1/2 hours.

The total hours that Cesar has used in practicing will be:

= 1 1/2 + 2 1/4 + 1 1/2

= 5 1/4

To get the number of hours that Cesar need to practice during the rest of the week in order to have his 10 total hours, we subtract 5 1/4 hours from 10 hours. This will be:

= 10 - 5 1/4

= 4 3/4 hours.

Cesar needs to practice for 4 3/4 hours more.

5 0
3 years ago
Solve the differential equation. y' + 5xey = 0.
Gwar [14]

Answer:

The solution is     y = - ln(\frac{5}{2}x^{2} + C)

Step-by-step explanation:

To solve the differential equation, we will find y

From the given equation, y' + 5xey = 0.

That is, y' + 5xe^{y} = 0

This can be written as

\frac{dy}{dx} + 5xe^{y} = 0

Then,

\frac{dy}{dx} = - 5xe^{y}

\frac{dy}{e^{y}}   = - 5x dx

Then, we integrate both sides

\int {\frac{dy}{e^{y}}}  =\int {- 5x dx}

\int {e^{-y}dy }}  =\int {- 5x dx}

Then,

-e^{-y} = -\frac{5}{2}x^{2} + C

e^{-y} = \frac{5}{2}x^{2} + C

Then,

ln(e^{-y}) = ln(\frac{5}{2}x^{2} + C)

Then,

-y = ln(\frac{5}{2}x^{2} + C)

Hence,

y = - ln(\frac{5}{2}x^{2} + C)

8 0
2 years ago
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