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prisoha [69]
3 years ago
9

8pq + 20qr – 16s factorise plz

Mathematics
1 answer:
Dafna1 [17]3 years ago
8 0

Answer:

4(2qp + 5qr - 4s)

Step-by-step explanation:

8pq + 20qr – 16s

we can use 4 as a common factor between these numbers.

8/4 = 2

20/4 = 5

16/4 = 4

so

4(2qp + 5qr - 4s) is the most factorized answer we can get

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Write an equation for the quadratic graphed below: x-intercepts: (-1,0) and (4,0); y-intercept: (0,1)
Luda [366]

Answer:

y = (1/4)x² - (5/4)x + 1

Step-by-step explanation:

The x-intercepts of the quadratic equation are simply it's roots.

Thus, we have;

(x + 1) = 0 and (x - 4) = 0

Now, formula for quadratic equation is;

y = ax² + bx + c

Where c is the y intercept.

At y-intercept: (0,1), we have;

At (-1,0), thus;

0 = a(1²) + b(1) + 1

a + b = -1 - - - (1)

At (4,0), thus;

0 = a(4²) + b(4) + 1

16a + 4b = -1

Divide both sides by 4 to get;

4a + b = -1/4 - - - (2)

From eq 1, b = -1 - a

Thus;

4a + (-1 - a) = -1/4

4a - 1 - a = -1/4

3a - 1 = -1/4

3a = 1 - 1/4

3a = 3/4

a = 1/4

b = -1 - 1/4

b = -5/4

Thus;

y = (1/4)x² - (5/4)x + 1

6 0
2 years ago
A triangle with sides 11m , 13m and 18m is a right triangle.<br> ​<br> A<br> True<br> B<br> False
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\bold{\huge{\pink{\underline{ Solution}}}}

\bold{\underline{ Given :-}}

  • <u>A </u><u>triangle </u><u>with </u><u>sides </u><u>11m</u><u>, </u><u> </u><u>13m </u><u>and </u><u>18m</u>

\bold{\underline{ To \: Find :-}}

  • <u>We</u><u> </u><u>have </u><u>to </u><u>check </u><u>it </u><u>whether </u><u>it </u><u>is </u><u>right </u><u>angled </u><u>triangle </u><u>or </u><u>not</u><u>? </u>

\bold{\underline{ Let's \: Begin :-}}

\sf{\red{ In \:right \:angled\ : triangle, }}

According to the Pythagoras theorem, The sum of the squares of perpendicular height and the square of the base of the triangle is equal to the square of hypotenuse that is sum of the squares of two small sides equal to the square of longest side of the triangle.

<u>We </u><u>imply</u><u> </u><u>it </u><u>in </u><u>the </u><u>given </u><u>triangle </u><u>,</u>

\sf{\red{ ( Perpendicular)² + (Base)² = (Hypotenuse)²}}

\sf{(AB)² + (BC)² = (AC)²}

\sf{ (11)² + (13)² = (18)² }

\sf{ 121 + 169 ≠  324 }

\sf{ 290 ≠ 324  }

<u>From </u><u>Above </u><u>we </u><u>can </u><u>conclude </u><u>that</u><u>, </u>

The sum of the squares of two small sides that is perpendicular height and base is not equal to the square of longest side that is Hypotenuse

\sf{\blue{ Hence,\: Your\: answer \:is \:false }}

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e position of a submarine relative to the water surface was −32 1/4 feet. A downward navigational maneuver increased its depth b
swat32

Answer:-47 3/4


Step-by-step explanation:

The correct answer would be -47 3/4  


You would use the equation


-32 1/4 - 15 1/2 = -47 3/4


to find your answer

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Answer:

Step-by-step explanation:

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