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Ann [662]
3 years ago
14

What is the degree of the equation?

Mathematics
1 answer:
Kisachek [45]3 years ago
5 0

Answer:

Degree: x= -1, \frac{8}{5}

leading coefficient: 5

Step-by-step explanation:

hope this helps!!!

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If 7/12x-1/3=1/2+3/8 what is the value of x
telo118 [61]

\dfrac{7}{12}x-\dfrac{1}{3}=\dfrac{1}{2}+\dfrac{3}{8}\\\\LCM(12,\ 3,\ 2,\ 8)=24\\\\\text{multiply both sides by 24}\\\\24\cdot\dfrac{7}{12}x-12\cdot\dfrac{1}{3}=12\cdot\dfrac{1}{2}+24\cdot\dfrac{3}{8}\\\\2\cdot7x-8=12+3\cdot3\\\\14x-8=12+9\\\\14x-8=21\qquad|\text{add 8 to both sides}\\\\14x=29\qquad|\text{divide both sides by 14}\\\\x=\dfrac{29}{14}

3 0
3 years ago
20. How many minutes would it take a train to travel
vladimir1956 [14]

Answer:

45

Step-by-step explanation:

Every 15 minutes, 30 kilometers are passed by.

15/60 = 1/4 hour

30/120 = 1/4 speed

3 0
3 years ago
Read 2 more answers
What is the length in feet of the base of the triangle?
worty [1.4K]

\bf \textit{area of a triangle}\\\\ A=\cfrac{1}{2}bh~~ \begin{cases} b=base\\ h=height\\[-0.5em] \hrulefill\\ A=\frac{2}{5}\\ h=\frac{6}{5} \end{cases}\implies \cfrac{2}{5}=\cfrac{1}{2}b\left( \cfrac{6}{5} \right)\implies \cfrac{2}{5}=\cfrac{6b}{10}\implies \cfrac{2}{5}=\cfrac{3b}{5} \\\\\\ 10=15b\implies \stackrel{\textit{dividing both sides by 15}}{\cfrac{10}{15}=\cfrac{15b}{15}}\implies \cfrac{10}{15}=b\implies \cfrac{2}{3}=b

7 0
4 years ago
The circle passes through the point (-1.-6). What is its radius?
Mice21 [21]

Answer:

Step-by-step explanation:

Not enough information.

7 0
3 years ago
In the right triangle shown, \angle B = 60^\circ∠B=60
Liono4ka [1.6K]

Answer:

Part 1) AC=6\sqrt{3}\ units

Part 2) AC=12\sqrt{3}\ units

Step-by-step explanation:

I will analyze two problems

see the attached figure to better understand the problem

Problem 1

The hypotenuse is the segment AB  and the right angle is C

we know that

In the right triangle ABC

sin(B)=\frac{AC}{AB} ---> by SOH (opposite side divided by the hypotenuse)

substitute the given values

sin(60^o)=\frac{AC}{12}

Remember that

sin(60^o)=\frac{\sqrt{3}}{2}

substitute

\frac{\sqrt{3}}{2}=\frac{AC}{12}

AC=6\sqrt{3}\ units

Problem 2

The hypotenuse is the segment BC  and the right angle is A

we know that

In the right triangle ABC

tan(B)=\frac{AC}{AB} ---> by TOA (opposite side divided by the adjacent side)

substitute the given values

tan(60^o)=\frac{AC}{12}

Remember that

tan(60^o)=\sqrt{3}

substitute

\sqrt{3}=\frac{AC}{12}

AC=12\sqrt{3}\ units

5 0
3 years ago
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