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ra1l [238]
3 years ago
5

HELP I WILL GIVE ALl MY POINTS NEXT QUESTION TO TH PERSON WHO ANSWERS RIGHT HELPP

Mathematics
1 answer:
djverab [1.8K]3 years ago
7 0

Answer:

Mixed Fraction -

1  \frac{5}{8}

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Solve a = b - 4/ c for b.
ahrayia [7]

Answer:

ac + 4 = b

Step-by-step explanation:

Step 1: Write equation

a = (b - 4)/c

Step 2: Multiply both sides by <em>c</em>

ac = b - 4

Step 3: Add 4 on both sides

ac + 4 = b

6 0
3 years ago
Read 2 more answers
A right triangle has a side with length 12 in and a hypotenuse with length 20 in. find the length of the missing leg. (round to
OLEGan [10]
Let the unknown side be "z" inches
Through pythagoras:
{12}^{2}  +  {z}^{2}  =  {20}^{2}
Because the square of the two sides add up to the square of the hypotenuse in a right triangle.

This means that
{z }^{2}  = {20}^{2}  -  {12}^{2}
{z}^{2}  = 400 - 144 = 256
z =  \sqrt{256}  = 16
So the missing side is 16 inches

Hope this helped
6 0
4 years ago
What is 4+2 x { 16/ (5+3)} 2
Dahasolnce [82]
The answer for you problem is 6
4 0
3 years ago
Read 2 more answers
Triangle P Q R is shown. Angle Q R P is a right angle. Angle R P Q is 30 degrees and angle P Q R is 60 degrees. Given right tria
mote1985 [20]

Answer:

The correct option is option (c).

Sin \ P= \frac {RQ}{PQ}

Step-by-step explanation:

Right angled triangle:

  • One angle must be 90° and other two angles are acute angle.
  • The hypotenuses is the longest side of the triangle and opposite right angle.
  • It follows the Pythagorean Theorem.

Given that,

∠QRP= 90°, ∠RPQ= 30°, ∠PQR = 60°

we know that,

sin \theta =\frac{Opposite }{Hypotenuse}

for sin P , the opposite is QR.

The hypotenuse is PQ.

Therefore,

Sin \ P= \frac {RQ}{PQ}

5 0
3 years ago
Read 2 more answers
What must be true for lines a and b to be parallel lines? Check all that apply
iren [92.7K]

the question does not present the options, but this does not interfere with the resolution


we know that

if a and b are parallel lines

so


1) m∠2=58°------> by corresponding angles


2) m∠1=4x-10------> by alternate exterior angles


3) [m∠2+m∠(3x-1)]+m ∠1=180°------> by supplementary angles

58+(3x-1)+4x-10=180

7x=180-47

7x=133

x=19°


4) angle (4x-10)=-4*19-10-------> 66°


5) angle 3x-1=3*19-1-------> 56°

3 0
3 years ago
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