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2 years ago

Whats the best grunge rock band?

Mathematics

2 answers:

sleet_krkn [62]2 years ago

3 0

Answer:

Matallica

Step-by-step explanation:

It's my personal favorite but also Nervana....

Y_Kistochka [10]2 years ago

3 0

Answer: Nirvana

Step-by-step explanation: hoped this helped :)

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Determine whether the two triangles are similar. If they are similar, write the singularity statement

Keith_Richards [23]

Answer:

Step-by-step explanation:

6 0

3 years ago

What are the x-intercept and y-intercept of the graph of y=13x−6 ?

gtnhenbr [62]

To find the x-int., let y = 0 and solve for x: 13x = 6, so x = 6/13: (6/13, 0)

To find the y-int., let x =0 and read off the y-int: (0, -6)

7 0

2 years ago

Read 2 more answers

Milan drove 715 miles in 11 hours. At the same rate, how many miles would he drive in 9 hours?

Gwar [14]

Answer:

585 miles

Step-by-step explanation:

Basically since he drove 715 miles in 11 hours you would have to divide the 2 numbers then you get 63.

63 is the rate so then you multiply by 9 hours then that's your miles.

585 miles

7 0

2 years ago

A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra

Harman [31]

Answer:

A.

The student made an error in step 3 because a is positive in Quadrant IV; therefore,

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

Step-by-step explanation:

Given

$P\ (a,b)$

$r = \± \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}$

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

$r = \sqrt{(a)^2 + (b)^2}$

Since a belongs to the x axis and b belongs to the y axis;

$cos\theta$ is calculated as thus

$cos\theta = \frac{a}{r}$

Substitute $r = \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}$

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}}$

Rationalize the denominator

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}$

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

So, from the list of given options;

The student's mistake is that a is positive in quadrant iv and his error is in step 3

3 0

2 years ago

Solve each equation for p. P+3/m=-1

Otrada [13]

Answer:

P= -m-3

I hope this helps!

3 0

3 years ago