All categories

3 years ago

Graph the line. Y= 4/3 x

Mathematics

1 answer:

s344n2d4d5 [400]3 years ago

8 0

Answer:

hope this helps :)

Step-by-step explanation:

You might be interested in

Can someone help find the area of this?

GuDViN [60]

You must do 1/2 times base times height

6 0

3 years ago

If f(x) = 3^2 + 10x and g(x) = 2x-4, find (f-g)(x).

Marat540 [252]

What we know:

$f(x)=3^2+10x\\g(x)=2x-4$

(Also, $3^2=9$ )

What we need to solve:

$(f-g)(x)$ (This is equal to $f(x)-g(x)$ )

How you have to subtract $f(x)$ with $g(x)$ to get:

$(f-g)(x) = f(x)-g(x) = 9+10x+2x-4 = 12x+5$

Thus, the answer would be:

$12x+5$

3 0

3 years ago

4. Match these equation balancing steps with the description of what was done in each step.

Archy [21]

Answer:

step 1 - C.

step 2 - A.

step 3 - B.

6 0

3 years ago

If 8 people do a job in 40 hours. How many people need to do the job in 32 hours?

bija089 [108]

Answer:

6 people

Step-by-step explanation:

8/40 = x/32

multiply both sides by 32: 8(32)/40 = x

x = 256/40

= 6.4

≈ 6

8 0

3 years ago

Find the exact value of the trigonometric expression.

Dmitrij [34]

$\bf \textit{Half-Angle Identities}
\\\\
sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}}
\qquad 
cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}\\\\
-------------------------------\\\\
\cfrac{1}{12}\cdot 2\implies \cfrac{1}{6}\qquad therefore\qquad \cfrac{\pi }{12}\cdot 2\implies \cfrac{\pi }{6}~thus~ \cfrac{\quad \frac{\pi }{6}\quad }{2}\implies \cfrac{\pi }{12}$

$\bf sec\left( \frac{\pi }{12} \right)\implies sec\left( \cfrac{\frac{\pi }{6}}{2} \right)\implies \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\impliedby \textit{now, let's do the bottom}
\\\\\\
cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+cos\left( \frac{\pi }{6} \right)}{2}}\implies \pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies \pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}}$

$\bf \pm \sqrt{\cfrac{2+\sqrt{3}}{4}}\implies \pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\implies \pm \cfrac{\sqrt{2+\sqrt{3}}}{2}
\\\\\\
therefore\qquad \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\implies \cfrac{2}{\sqrt{2+\sqrt{3}}}$

which simplifies thus far to

$\bf \cfrac{2}{\sqrt{2+\sqrt{3}}}\cdot \cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\cdot \stackrel{conjugate}{\cfrac{2-\sqrt{3}}{2-\sqrt{3}}}
\\\\\\
\cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{2^2-(\sqrt{3})^2}\implies \cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{1}$

$\bf 2\sqrt{2+\sqrt{3}}(2-\sqrt{3})\impliedby \stackrel{\textit{keep in mind that}}{(2-\sqrt{3})=(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}})}
\\\\\\
2\sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}
\\\\\\
2\sqrt{(2+\sqrt{3})(2-\sqrt{3})\cdot (2-\sqrt{3})}
\\\\\\
2\sqrt{[2^2-(\sqrt{3})^2]\cdot (2-\sqrt{3})}\implies 2\sqrt{1(2-\sqrt{3})}
\\\\\\
2\sqrt{2-\sqrt{3}}$

7 0

4 years ago