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Musya8 [376]
3 years ago
14

Can you help me with these 4 problems?

Mathematics
1 answer:
adoni [48]3 years ago
3 0
What are the questions attach a pic plz
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Subtract. Write your answer in simplest form.<br><br> 3 1/8 - 7/12<br><br> 2<br> 2<br> 3<br> 3
Agata [3.3K]
2 13/24
i hope this helped
8 0
3 years ago
How do you do inequalities
Alexandra [31]

Answer:

Summary Many simple inequalities can be solved by adding, subtracting, multiplying or dividing both sides until you are left with the variable on its own. But these things will change direction of the inequality: Multiplying or dividing both sides by a negative number. Swapping left and right hand sides

Step-by-step explanation:

6 0
2 years ago
Solve for x with step by step explanation
HACTEHA [7]

0.85x-100<80

collect like terms

0.85x<80+100

0.85x<180

divide by 0.85

x<211.8

6 0
3 years ago
How many solutions does the system have?
marishachu [46]

Answer:

{x,y} = {-2,2}

Step-by-step explanation:

Solve equation [2] for the variable  y  

 

  7y = 7x + 28

  y = x + 4

Plug this in for variable  y  in equation [1]

-8x + 4•(x +4) = 24

-4x = 8

Solve equation [1] for the variable  x  

4x = - 8  

x = - 2

By now we know this much :

   x = -2

   y = x+4

Use the  x  value to solve for  y  

   y = (-2)+4 = 2

Solution :

{x,y} = {-2,2}

and this intersects at only one point so your answer is A

3 0
3 years ago
From the side view, a gymnastics mat forms a right triangle with other angles measuring 60° and 30°. The gymnastics mat extends
valkas [14]

Answer: The mat is 4.33 ft high off the ground.

Step-by-step explanation:

Since we have given that

Angle of elevation with the first triangle = 30°

Angle of elevation with the second triangle = 60°

Length at which gymnastics mat extends across the floor = 5 feet

so, As shown in the figure:

We need to find the height of the mat off the ground.

If CD = 5 ft,

Let,  AB = y, DC = x.

In Δ ABC,

\tan 60^\circ=\frac{AB}{BC}\\\\\frac{\sqrt{3}}{2}=\frac{y}{x}\\\\x=\frac{y}{\sqrt{3}}

Similarly, in Δ ACD,

\tan 30^\circ=\frac{AB}{BD}\\\\\frac{1}{\sqrt{3}}=\frac{y}{x+5}\\\\\frac{1}{\sqrt{3}}=\frac{y}{\frac{y}{\sqrt{3}}+5}\\\\\frac{1}{\sqrt{3}}=\frac{y\sqrt{3}}{y+5\sqrt{3}}\\\\3y=y+5\sqrt{3}\\\\2y=5\sqrt{3}\\\\y=\frac{5\sqrt{3}}{2}\\\\y=4.33\ ft

Hence, the mat is 4.33 ft high off the ground.

5 0
3 years ago
Read 2 more answers
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