All categories

3 years ago

(3,7) Quadrant where the point is located

Mathematics

1 answer:

AleksandrR [38]3 years ago

6 0

Since both values are positive, it is in quadrant 1

You might be interested in

Which shows the unsimplified solution for x^2 + 8x + 6 = 0?

irinina [24]

<u /> $x^{2} + 8x + 6 = 0 \\x = \frac{-(8) +/- \sqrt{(8)^{2} - 4(1)(6)}}{2(1)} \\x = \frac{-8 +/- \sqrt{64 - 24}}{2} \\x = \frac{-8 +/- \sqrt{40}}{2} \\x = \frac{-8 +/- 2\sqrt{10}}{2} \\x = -4 +/- \sqrt{10}$
$x = -4 + \sqrt{10}$ $x = -4 - \sqrt{10}$

It is equal to D.

6 0

3 years ago

Explain how the expression 8.3√2+7.9√2 can be simplified using the distributive property.

IceJOKER [234]

Answer:

$\sqrt{2}(16.2)$

Step-by-step explanation:

The distributive property says that the answer of multiplying two numbers is the same as multiplying the first number by the sum or subtraction of two numbers that result in the second number.

What we have to do is to simpligy the ecuation, this mean, making it shorter.

First we take out the common term of the ecuation: $\sqrt{2}$ .

By distributive property $8.3\sqrt{2} +7.9\sqrt{2} =\sqrt{2}(8.3+7.9)$

then $\sqrt{2} (16.2)$ .

3 0

3 years ago

(0) 14 -7 0<br> | () 14 - 7 0

lyudmila [28]

Answer:

What is the question.

5 0

4 years ago

Prism A and Prism B have identical bases. The height of prism B is twice the height of prism A Determine the relationship betwee

Semmy [17]

Answer:

it is B, the volume of the prism B is 2 times the volume of Prism A

5 0

3 years ago

I don’t know how to solve this

Lady_Fox [76]

Answer:

$\theta =2\pi k,\ \ k\in Z\ \\\text{or}\ \\\theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z$

Step-by-step explanation:

Given:

$\cos \theta-\sqrt{3}\sin \theta=1$

Divide this equation by 2:

$\dfrac{1}{2}\cos \theta-\dfrac{\sqrt{3}}{2}\sin \theta=\dfrac{1}{2}$

Note that

$\cos \dfrac{\pi }{3}=\dfrac{1}{2}\\ \\\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$

So, the previous equation is

$\cos \dfrac{\pi}{3}\cdot \cos \theta-\sin \dfrac{\pi}{3}\cdot \sin \theta=\dfrac{1}{2}$

Remind that

$\cos x\cos y-\sin x\sin y=\cos (x+y),$

then

$\cos \left(\dfrac{\pi}{3}+\theta\right)=\dfrac{1}{2}$

The solution of this equation is

$\dfrac{\pi}{3}+\theta=\pm \arccos \dfrac{1}{2}+2\pi k,\ \ k\in Z\\ \\\dfrac{\pi}{3}+\theta=\pm \dfrac{\pi}{3}+2\pi k,\ \ k\in Z\\ \\\theta =2\pi k,\ \ k\in Z\ \text{or}\ \theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z$

8 0

3 years ago