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2 years ago

In need help with this finding function

Mathematics

1 answer:

hjlf2 years ago

4 0

Is there a picture? I can’t see one

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Graph the following linear function.<br> f(x) = 1/6x+3

Black_prince [1.1K]

Y-axis is 3
x-axis is -18

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2 years ago

If you want to know the probability of spinning a spinner in a game that has 8 equal-sized

aleksandr82 [10.1K]

the answer is 1/8. Because you divide how many pieces there are by that same number.

7 0

2 years ago

Read 2 more answers

Find how many quarts of 6% butterfat milk and 1% butterfat milk should be mixed to yield 75 quarts of 3% butterfat milk.

IgorC [24]

Answer:

30 gallons of 6% milk, and 45 gallons of 1% milk.

Step-by-step explanation:

x=quarts of 6% butterfat

<h2>First, write an equation.</h2>

0.06x+0.01(75-x)=0.03*75

<h2>Multiply both sides by 100.</h2>

0.06x*100+0.01*100(75-x)=100*0.03*75

<h2>Simplify</h2>

6x+75-x=3*75

6x+75-x=225

<h2>Solve</h2>

5x=225-75=150

5x=150

x=30

75-30=45

So, we need 30 gallons of 6% milk, and 45 gallons of 1% milk.

8 0

3 years ago

Read 2 more answers

Find X using the image above

Andrej [43]

hope it helps you friend....!!

Please give Brainlist if it helps...

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2 years ago

General solutions of sin(x-90)+cos(x+270)=-1<br> {both 90 and 270 are in degrees}

mixer [17]

Answer:

$\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

Step-by-step explanation:

Given:

$\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1$

First, note that

$\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x$

So, the equation is

$-\cos x+\sin x= -1$

Multiply this equation by $\frac{\sqrt{2}}{2}:$

$-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}$

The general solution is

$x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

4 0

3 years ago