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AleksandrR [38]

2 years ago

5

What is the minimum value for the function shown in the graph? Enter your answer in the box.

1 answer:

ch4aika [34]2 years ago

8 0

Check the picture below.

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zzz [600]

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3 years ago

A hose fills an empty fish tank with 24 gallons of water in 8 minutes. At that same rate, how many minutes will it take the hose

kolezko [41]

Answer:

$\frac{8}{24} = \frac{x}{60}$

24x = 8*60

24x = 480

x = 20

Step-by-step explanation:

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3 years ago

Rewrite the expression in the form k - x^n.

JulsSmile [24]

Answer:

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6 0

3 years ago

5a +4b+c-3a-c+4b simplified

MAVERICK [17]

Answer:

2a + 8b

Step-by-step explanation:

4 0

3 years ago

Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3> (1 point)

sesenic [268]

Angle between $u = -5i-4j , v=-4i-3j$ is $x =0$ ° .

<u>Step-by-step explanation:</u>

We have , two vectors u = <-5, -4>, v = <-4, -3> or , $u = -5i-4j , v=-4i-3j$

We need to find angle between these two vectors . Let's find out:

We know that dot product of two vectors is defined as :

$u.v =|u|(|v|)cosx$ , where x is angle between u & v !

⇒ $u.v =|u|(|v|)cosx$

⇒ $cosx =\frac{u.v}{|u|(|v|)}$

Now , $u.v = (-5i-4j)(-4i-3j)$

⇒ $u.v = (-5i-4j)(-4i)-(-5i-4j)(3j)$

⇒ $u.v = 20+12$ { $i(j) = j(i) =0$ }

⇒ $u.v = 32$

Now , Modulus of any vector $r = xi+yj$ is $|r| = \sqrt{x^{2}+y^{2}}$ So ,

$|u| = \sqrt{(-5)^{2}+(-4)^{2}} = \sqrt{25+16} = \sqrt{41} \\\\|v| = \sqrt{(-4)^{2}+(-3)^{2}} = \sqrt{16+9} = \sqrt{25} = 5$

Putting all these values in equation $cosx =\frac{u.v}{|u|(|v|)}$ we get:

⇒ $cosx =\frac{32}{5(\sqrt{41})}$

⇒ $cos^{-1}(cosx) =cos^{-1}(\frac{32}{5(6.4)})$

⇒ $x =cos^{-1}(1)$ { $cos0 = 1$ }

⇒ $x =0$ °

Therefore , Angle between $u = -5i-4j , v=-4i-3j$ is $x =0$ ° .

8 0

3 years ago