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3 years ago

What is the quotient 8 x 4 − 12 x 2 − 16 x 4 x , where x ≠ 0 ?

Mathematics

1 answer:

Nat2105 [25]3 years ago

5 0

Answer:

that = 8 and if you want to see more you can go on

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Rewrite the following expression x^9/7

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Answer:

$\sqrt[7]{x^9}$

Step-by-step explanation:

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In the figure, TP and TS are opposite rays. TQ bisects < RTP.

Ket [755]

Answer:

search-icon-image

Class 9

>>Maths

>>Quadrilaterals

>>Quadrilaterals and Their Various Types

>>In Fig. 6.43, if PQ PS, PQ∥ SR, SQR = 2

Question

Bookmark

In Fig. 6.43, if PQ⊥PS,PQ∥SR,∠SQR=28

and ∠QRT=65

, then find the values of x and y.

463685

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Medium

Solution

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Given, PQ⊥PS,PQ∥SR,∠SQR=28

∘

,∠QRT=65

∘

According to the question,

x+∠SQR=∠QRT (Alternate angles as QR is transversal.)

⇒x+28

∘

=65

∘

⇒x=37

∘

Also ∠QSR=x

⇒∠QSR=37

∘

Also ∠QRS+∠QRT=180

∘

(Linear pair)

⇒∠QRS+65

∘

=180

∘

⇒∠QRS=115

∘

Now, ∠P+∠Q+∠R+∠S=360

∘

(Sum of the angles in a quadrilateral.)

⇒90

∘

+65

∘

+115

∘

+∠S=360

∘

⇒270

∘

+y+∠QSR=360

∘

⇒270

∘

+y+37

∘

=360

∘

⇒307

∘

+y=360

∘

⇒y=53

∘

Step-by-step explanation:

please mark me as brainlist please

6 0

2 years ago

Plz help as quick as possible!

kompoz [17]

average =65%

average = (60+70)/(100+100)*100

= 130/200*100

= 65%

6 0

3 years ago

Read 2 more answers

(ii) Hence solve the following system of simultaneous equations:

gladu [14]

Answer:

Hi how are you doing today Jasmine

7 0

3 years ago

Which equations represent the line that is perpendicular to the line 5x − 2y = −6 and passes through the point (5, −4)? Check al

Alla [95]

Let's rewrite each equation in the Slope-Intercept Form of the Equation of a Line. First, let's start with the main equation:

$\bullet \ 5x-2y=-6 \therefore y=\frac{5}{2}x+3$

Then, our options are the following:

$A) \ y = -\frac{2}{5}x-2 \\ \\ B) \ 2x+5y=-10 \therefore y=-\frac{2}{5}x-2 \\ \\ C) \ 2x-5y=-10 \therefore y=\frac{2}{5}x+2 \\ \\ D) \ y+4=-\frac{2}{5}(x-5) \therefore y=-\frac{2}{5}x-2 \\ \\ E) \ y-4=\frac{5}{2}(x + 5) \therefore y=\frac{5}{2}x+\frac{33}{2}$

For two perpendicular lines it is true that the product of its slopes is:

$m_{1}m_{2}=-1$

$m_{1}m_{2}=-1 \\ \\ m_{1} \ is \ the \ slope \ of \ y=\frac{5}{2}x+3, \ that \ is, \ m_{1}=\frac{5}{2} \\ \\ Then, \ the \ slope \ of \ a \ perpendicular \ line \ is: \\ \\ m_{2}=-\frac{2}{5}$

According to this, only A) B) and D) might be the perpendicular lines we are looking for. Notice that these lines are the same. The other condition is that the line must pass through the point (5, -4). By substituting this point in the equation, we have:

$y = -\frac{2}{5}x-2 \\ \\ -4=-\frac{2}{5}(5)-2 \\ \\ -4=-2-2 \\ \\ \boxed{-4=-4} \ True!$

Finally, the right answer are:

$A) \ y = -\frac{2}{5}x-2 \\ \\ B) \ 2x+5y=-10 \\ \\ D) \ y+4=-\frac{2}{5}(x-5)$

8 0

3 years ago

Read 2 more answers