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2 years ago

What is the answer to four thirds divides by five thirds?

1 answer:

3 0

Copy Dot Flip.
so Copy the first fraction 4/3
write a Dot • or the multiplication, so 4/3 ×
Flip the last fraction 3/5
So...
4/3 × 3/5 = 12/15 = 4/5

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Write an equation for the line perpendicular to 4x - 12y = 2 and passing through point (10, -1)

MrMuchimi

Answer:

3x+y=-31

Step-by-step explanation:

$4x - 12y = 2\\12y = 4x-2\\y=\frac{x}{3}-\frac{1}{6}$

Slope of perpendicular line is -3.

$y=-3x+b$

Sub in the point (10, -1)

$-1=-3(10)+b\\-1=30+b\\b=-31\\y=-3x-31\\3x+y=-31$

8 0

2 years ago

Caleb wants to work as a dog walker. One client offers him $10 per day to walk his dog Puff, and wants Caleb to walk Puff every

aksik [14]

Answer:

he will earn more from walking Puff

Step-by-step explanation:

if he is walking Puff for ten dollars 7 days a week, thats 70 dollars.

the other dog would get him 60 dollars a week.

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2 years ago

Please help! Number 9) picture is included! Determine the base of the rectangle. (10 points) please help as soon as possible! :)

Juli2301 [7.4K]

The answer is 128m^2/8m equals 16m

3 0

2 years ago

Jamie takes at random a counter from the bag and records the colour. Red : blue : green = 1:3:7 Jamie does this a number of time

Sonbull [250]

Answer: There are 249 times Jamie takes a counter from the bag.

Explanation:

Since we have given that

Ratio of colour of red to blue to green is 1:3:7.

Number of blue counters = 68

Let the total number of times Jamie does from the bag be x.

So, According to question,

$\frac{3}{11}\times x=68\\\\x=\frac{11}{3}\times 68\\\\x=249.33\\\\=249\ approx.$

Hence, there are 249 times Jamie takes a counter from the bag.

8 0

2 years ago

Use the Law of Sines to find the measure of angle J to the nearest degree.

rjkz [21]

$\bf \textit{Law of sines}
\\ \quad \\
\cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\\\\
-----------------------------\\\\
\cfrac{sin(J)}{9.1}=\cfrac{sin(97^o)}{11}\implies sin(J)=\cfrac{9.1\cdot sin(97^o)}{11}
\\\\\\
\textit{now taking }sin^{-1}\textit{ to both sides}
\\\\\\
sin^{-1}\left[ sin(J) \right]=sin^{-1}\left( \cfrac{9.1\cdot sin(97^o)}{11} \right)
\\\\\\
\measuredangle J=sin^{-1}\left( \cfrac{9.1\cdot sin(97^o)}{11} \right)$

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3 years ago