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Katen [24]
2 years ago
14

Find the x - and y -intercepts of the graph of the linear equation 3x+6y=24 .

Mathematics
2 answers:
Vedmedyk [2.9K]2 years ago
6 0

Answer:

x -intercept is 8

y -intercept is 4

Step-by-step explanation:

SOLUTION:

(x -intercept)

3x+6y=24

3x+6×0=24

3x+0= 24

(3x/3= 24/3) divide both sides by 3

x=8

____________

(y -intercept)

3x+6y=24

3×0+6y=24

(6y/6=24/6) divide both sides by 6

y=4

•| HOPE IT HELPS |•

True [87]2 years ago
5 0

Answer:

x-intercept(s):

(8,0)

y-intercept(s):

(0,4)

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16, 17, 18, 19, 20, 21, 22, 23, 24, and 25

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<img src="https://tex.z-dn.net/?f=%282%20%5Csqrt%7B3%20%2B%203%20%5Csqrt%7B2%29%20%7B%3F%7D%5E%7B2%7D%20%7D%20%7D%20" id="TexFor
AlladinOne [14]

Answer:

\left(2\:\sqrt{3\:+\:3\:\sqrt{2}\:\:}\:\right)\:^2=4(3+3\sqrt{2})

Step-by-step explanation:

Considering the radical expression

\left(2\:\sqrt{3\:+\:3\:\sqrt{2}\:\:}\:\right)\:^2

​\mathrm{Apply\:exponent\:rule}:\quad \left(a\cdot \:b\right)^n=a^nb^n

=2^2\left(\sqrt{3+3\sqrt{2}}\right)^2.....[A]

Simplifying

\left(\sqrt{3+3\sqrt{2}}\right)^2

\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}=a^{\frac{1}{2}}

=\left(\left(3+3\sqrt{2}\right)^{\frac{1}{2}}\right)^2

\mathrm{Apply\:exponent\:rule}:\quad \left(a^b\right)^c=a^{bc}

=\left(3+3\sqrt{2}\right)^{\frac{1}{2}\cdot \:2}

=3+3\sqrt{2}          As \frac{1}{2}\cdot \:2=1

So, putting 3+3\sqrt{2} into Equation [A]

=2^2\left(\sqrt{3+3\sqrt{2}}\right)^2.....[A]

As

\left(\sqrt{3+3\sqrt{2}}\right)^2=3+3\sqrt{2}

So, Equation [A] becomes

=2^{2}(3+3\sqrt{2})

=4(3+3\sqrt{2})

Therefore,

\left(2\:\sqrt{3\:+\:3\:\sqrt{2}\:\:}\:\right)\:^2=4(3+3\sqrt{2})

Keywords: radical expression

Learn more radical expression from brainly.com/question/11624221

#learnwithBrainly

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