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3 years ago

Help please for homework

Mathematics

1 answer:

krek1111 [17]3 years ago

5 0

4 + 3m and 2x + 3y. Sum means addition

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30 points! Find the angles formed by two intersecting lines, if the sum of three of the four angles is equal to 270°.

nydimaria [60]

Each angle is 90 degrees because 90 times 3 is 270.

8 0

3 years ago

Help

kobusy [5.1K]

The equation of the line is y - 3 = 5(x - 8) ⇒ c

Step-by-step explanation:

The point-slope form of the linear equation is $y-y_{1}=m(x-x_{1})$ , where:

m is the slope of the line
$(x_{1},y_{1})$ is a point on the line

∵ The line passes through point (8 , 3)

∴ $x_{1}$ = 8 and $y_{1}$ = 3

∵ The slope of the line is 5

∴ m = 5

- Substitute these values in the form of the equation below

∵ $y-y_{1}=m(x-x_{1})$

∴ y - 3 = 5(x - 8)

The equation of the line is y - 3 = 5(x - 8)

Learn more:

You can learn more about the linear equations in brainly.com/question/1284310

#LearnwithBrainly

3 0

3 years ago

What are the explicit equation and domain for a geometric sequence with a first term of 4 and a second term of −12? an = 4(−3)n

nadya68 [22]

The geometric sequence is given by:
an=ar^(n-1)
where:
a=first term
r=common ratio
n is the nth term
given that a=4, and second term is -12, then
r=-12/4=-3
hence the formula for this case will be:
an=4(-3)^(n-1)
where n≥1

5 0

3 years ago

Read 2 more answers

If f(x)=3^x + 10^x and g(x)=4^x-2,find(f-g)(x)

tresset_1 [31]

H gc d rxr V unimoi oubyctvuimubyv

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3 years ago

For what value of constant c is the function k(x) continuous at x = 0 if k =

nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x) is continuous, we take the limit of the parameter as follows:

$\mathbf{ \lim_{x \to 0^-} k(x) = \lim_{x \to 0^+} k(x) = 0 }$

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= c }$

Provided that:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }$

Using l'Hospital's rule:

$\mathbf{\implies \lim_{x \to 0} \ \ \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}= \lim_{x \to 0} sec \ x \ tan \ x = 0}$

Therefore:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}=0 }$

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

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5 0

2 years ago