Solve the following system of equations algebraically<br>3×-y=0<br>

Solve s = rθ for θ.<br><br> A) θ = rs <br> B) θ = r/s<br> C) θ = s/r<br> D) θ = 1/rs

meriva

Dividing both sides by r

=s/r

5 0

3 years ago

Read 2 more answers

2 (Picture) CONVERGENT AND DIVERGENT SERIES PLEASE HELP!!

Ivanshal [37]

Answer:

This series is divergent

F

Step-by-step explanation:

we are given a series

Firstly, we will find nth term

Numerator:

3, 4, 5,...

so, nth term will be

$a_n=n+3$

Denominator:

4,5,6,....

so, nth term will be

$b_n=n+4$

so, we can find it's nth term as

$c_n=\frac{n+3}{n+4}$

we can use divergent test

$\lim_{n \to \infty} c_n=\lim_{n \to \infty} \frac{n+3}{n+4}$

we can divide top and bottom by n

$\lim_{n \to \infty} c_n= \lim_{n \to \infty} \frac{n/n+3/n}{n/n+4/n}$

$\lim_{n \to \infty} c_n= \lim_{n \to \infty} \frac{1+3/n}{1+4/n}$

now, we can plug n=inf

$\lim_{n \to \infty} c_n= \frac{1+0}{1+0}$

$\lim_{n \to \infty} c_n=1$

Since, it is non-zero value

so, this series is divergent

8 0

3 years ago

line pass through the line (-3, -7.5)tahila determined that the equation of line n is y=0.5x explain the error tahila made while

Tasya [4]

The equation of the line will be (assuming the y-intercept equal to zero):

y = 2.5*x

<h3>What is Tahila's mistake?</h3>

We know that we have a linear equation of the form:

y = a*x + b

Such that we know that the line passes through (-3, -7.5), notice that the proposed equation is:

y = 0.5*x

If you evaluate that in -3, you get:

y = 0.5*-3 = -1.5

So this line does not pass through (-3, -7.5).

If we assume that b = 0 in the linear equation, then we can find the value of a as:

-7.5 = a*-3

a = 7.5/3 = 2.5

Then the linear equation is:

y = 2.5*x

If you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

4 0

2 years ago

Y=1/6x + 5 what is the answer

kykrilka [37]

X should equal 6y-30

3 0

3 years ago

Part 2. State what additional information is required in order to know that the triangles are congruent by ASA.

Helga [31]

<h3>Answer: Choice D. $\angle SQR \cong \angle XQR$ </h3>

The red angle markers show those two angles are congruent. That's one "A" of "ASA". The S refers to a congruent pair of sides. We don't have any tickmarks to indicate congruent pairs; however, we do know that QR = QR is a shared side that overlaps (reflexive theorem). So this is the "S" in "ASA".

The thing missing is the angle Q of the top triangle, and also of the bottom triangle as well. If we know those two angles are congruent, then we have enough info to use ASA. More specifically, if we know that $\angle SQR \cong \angle XQR$ , then we can use ASA.

One thing to notice is that the other answer choices involve side lengths and not angles. This implies that if A, B or C were one of the answers, then we would have something like SAS or SSS. But instead we want ASA. So we can immediately rule choices A,B, and C out.

7 0

3 years ago

Solve the following system of equations algebraically3×-y=0​

Solve the following system of equations algebraically3×-y=0