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4 years ago

Please help! What is -9/5 - (-9/5)? A. 0 B.2 C. 18/5 D. -18/5

Mathematics

2 answers:

Pie4 years ago

8 0

Answer:6

Step-by-step explanation:

Anettt [7]4 years ago

6 0

Answer: its A. 0 because you have to change the subtraction to adding then make the (-9/5) to a positive

-9/5+9/5 = 0

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The list below gives the maximum temperatures (in ° C), for each month of the year from January through December in Fairbanks. W

Triss [41]

-25, -23, -16, -13, -4, -2, +17, +15, +14, +7, +9

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3 0

3 years ago

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I need help thanks so much

Lerok [7]

Answer:

c. 1 $\frac{3}{8}$ inches

Step-by-step explanation:

Since the paper clip is past the 1, the answer will be 1 and ___.

the mark in the middle of the 1 and 2 is one half (1/2)

the mark in between the 1 and 1/2 is 1/4

the mark between 1/4 and 1/2 is 1/8

there are three (3) 1/8 marks where the paperclip is

3 * 1/8 = 3/8

so your answer is 1 3/8

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6 0

2 years ago

there are 6 red marbles, 4 blue marbles, n 15 green marbles in jar. if you reach in n randomly draw one, what is the probability

Arte-miy333 [17]

Answer:

6÷25

Step-by-step explanation:

Favourable outcomes are 6,and the total outcomes are

$6 + 4 + 15 = 25$

So the probability of getting fed marble is Favourable outcome ÷ total outcome=6÷25.

7 0

3 years ago

Solve r = cs/d for c

Nady [450]

$r=\frac{cs}{d}\\\\\frac{cs}{d}=r\ \ \ \ |multiply\ both\ sides\ by\ d\\\\cs=dr\ \ \ \ \ \ |divide\ both\ sides\ by\ s\\\\\boxed{c=\frac{dr}{s}}$

6 0

3 years ago

Two forces with magnitudes of 150 and 100 pounds act on an object at angles of 30° and 120°, respectively. Find the direction an

lyudmila [28]

<h2>Answer:</h2>

$Magnitude=180.27 \ lbf \\ \\ Direction=63.69 \ degrees$

<h2>Step-by-step explanation:</h2>

We have two forces as follows:

<u>First force:</u>

Magnitude: 150 pounds

Angle: 30°

<u>First force:</u>

Magnitude: 100 pounds

Angle: 120°

So the components can be found as follows:

$F_{1x}=150cos(30)=129.90 \ lbf\\F_{1y}=150sin(30) = 75 \ lbf \\ \\ F_{2x}=100cos(120)=-50 \ lbf\\F_{2y}=100sin(120) = 86.60 \ lbf$

So the components of the resultant force can be found by adding each component of the individual forces as follows:

$R_{x}=\Sigma F_{x} \\ R_{y}=\Sigma F_{y} \\ \\ R_{x}=129.90-50=79.90 \ lbf \\ R_{y}=75+86.60=161.6 \ lbf$

Finally, the magnitude and direction of the resultant force is:

$Magnitude \rightarrow R=\sqrt{R_{x}^2+R_{y}^2}=\sqrt{79.90^2+161.6^2}=180.27 \ lbf \\ \\ Direction \rightarrow \theta=tan^{-1}(\frac{R_{y}}{R_{x}})=tan^{-1}(\frac{161.6}{79.90})=63.69 \ degrees$

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3 years ago

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