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Elza [17]
3 years ago
8

Is this a true or false equation? -5(3-9) 4 = 5q - 11

Mathematics
1 answer:
tino4ka555 [31]3 years ago
8 0
<span>yes it is a true equation.. you can solve it to find q ...
 
-5(3-9) 4 = 5q - 11
</span><span>−5 *−6 * 4= 5q−11</span><span>120 = 5q−11
120 + 11 = 5q
131 = 5q
 q = 131/5 
in decimals q = </span><span> 26.2 ..
</span><span>
Hope it helps!!!!

</span><span>

</span>
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9/10-1/3-1/5 find the fraction
Zanzabum
Find a common multiple in between the denominators.
10 and 3 and 5 -> the least common multiple is 30.

Change all the denominators to 30 by multiplying the top and bottom by the same number.

9/10=27/30
1/3=10/30
1/5=6/30

Subtract!
27/39-10/30-6/30=11/30

The answer is 11/30

Hope this helps!



3 0
3 years ago
Look at the picture ​
Nikitich [7]

Answer:

c

Step-by-step explanation:

4 0
2 years ago
What is the surface area of the right prism below?
-Dominant- [34]

Answer:

120 sq. units

plus them all and add 100

3 0
2 years ago
Read 2 more answers
part 3. Find the coordinates of the vertices of the triangle after a reflection across the y-axis and then across the line y= 2.
garik1379 [7]

Answer:

  c.  S'(3, 1), T'(1, -1), U'(0, 1)

Step-by-step explanation:

Reflection across the y-axis negates the x-coordinate, so is equivalent to the transformation ...

  (x, y) ⇒ (-x, y)

Reflection across the horizontal line y=c is equivalent to the transformation ...

  (x, y) ⇒ (x, 2c-y)

So, the combined reflections are equivalent to the transformation ...

  (x, y) ⇒ (-x, 4 -y)

Then we have ...

  S(-3, 3) ⇒ S'(-(-3), 4-3) = S'(3, 1)

  T(-1, 5) ⇒ T'(-(-1), 4-5) = T'(1, -1)

  U(0, 3) ⇒ U'(-(0), 4-3) = U'(0, 1) . . . . matches choice C

7 0
3 years ago
1. tan a = cot(a + 10°)
Temka [501]

Answer:

a=40+90n

Step-by-step explanation:

Use a cofunction identity on the right hand side or left hand side...

So \tan(a)=\cot(90-a).

We have the equation:

\tan(a)=\cot(a+10)

Make the above replacement:

\cot(90-a)=\cot(a+10)

Since cotangent has period 180 degrees, we can also write this as:

\cot(90-a+180n)=\cot(a+10)

So solving the following will give us a set of solutions for a:

90-a+180n=a+10

Add a on both sides:

90+180n=2a+10

Subtract 10 on both sides:

80+180n=2a

Divide both sides by 2:

40+90n=a

Symmetric property:

a=40+90n

7 0
2 years ago
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