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2 years ago

13

Need helppppppppppppppppp

1 answer:

Fudgin [204]2 years ago

7 0

2x+3=-27
2x=-27-3
2x=-30
x=-15

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Help plz,BIG POINTS,also plz show me the steps as best as possible

Sidana [21]

Answer:

The correct answer is option 3.

Step-by-step explanation:

Given : ΔPQR, QM is altitude of the triangle

PM = 8

MR = 18

To find = QM

Solution :

PR = 8 + 18 = 26

Let, PQ = x , QR = y, QM = z

Applying Pythagoras Theorem in ΔPQR

$PQ^2+QR^2=PR^2$

$x^2+y^2=(26)^2$ ..[1]

Applying Pythagoras Theorem in ΔPQM

$PM^2+QM^2=PQ^2$

$8^2+z^2=(x)^2$ ..[2]

Applying Pythagoras Theorem in ΔQMR

$MR^2+QM^2=QR^2$

$18^2+z^2=(y)^2$ ..[3]

Putting values of $x^2$ and $y^2$ from [2] and [3 in [1].

$8^2+z^2+18^2+z^2=(26)^2$

$2z^2=(26)^2-(8)^2-(18)^2$

$2z^2=288$

$z^2=144$

z = ±12

z = 12 = QM ( ignoring negative value)

The length of QM is 12.

6 0

3 years ago

G=x-c/x. for x<br><br>how do yoy make this an equation for x? help

zzz [600]

1.
$\displaystyle \text{if }\quad G=x- \frac{C}{x} \\\\$

$x \neq 0$

Multiply both sides by x:

$\displaystyle Gx=x\bigg(x- \frac{C}{x}\bigg )\\\\Gx=x^2-C$

Subtract from both sides Gx:
$Gx-Gx=x^2-C-Gx\\x^2-Gx-C=0$

Then solve this equation:

$\displaystyle x^2-Gx-C=0\\\\x_{1,2}= \frac{G\pm \sqrt{G^2+4C} }{2}$

2.
$\displaystyle\text{if}\quad G= \frac{x-C}{x}$

$x \neq 0$

Multiply both sides by x:

$Gx=x-C$

Subtract from both sides Gx:

$x-C-Gx=0$

Then solve this equation:

$\displaystyle x-Gx-C=0\\x(1-G)=C\\x= \frac{C}{1-G}$

7 0

3 years ago

How do you solve the substitution of y=5x-3 and y=-6x-3

svlad2 [7]

5x-3=y 6x-3=y
x=1 x=1
51-3=y 61-3=y
51=5 61=6
5-3=y 6-3=y
2 3
y=2 y=3

8 0

3 years ago

Hi guys I need help<br><br><br><br><br><br>please<br><br><br><br><br><br><br>

Zanzabum

Answer:

saannn Wala Naman

Step-by-step explanation:

HAHAHAJAH BLABLA

THANKS SA POINTS

7 0

3 years ago

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Please answer this correctly

Crank

Answer:

Its absolutely cone , it has one triangle face seen from front , and a circular base

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3 years ago

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