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rusak2 [61]
3 years ago
5

Name a pair of lines that appear to be perpendicular

Mathematics
1 answer:
Ivenika [448]3 years ago
5 0
2 plus 2 is 4 minus 1 thats 3 quick maths
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6. Using the discriminant, determine the value of k that will give 1 solution (i.e. discriminant equals zero) y = kx²-4x + 4 ​
Gwar [14]

Answer:

k = 1

Step-by-step explanation:

<u>Discriminant</u>

b^2-4ac\quad\textsf{when }\:ax^2+bx+c=0

\textsf{when }\:b^2-4ac > 0 \implies \textsf{two real roots}

\textsf{when }\:b^2-4ac=0 \implies \textsf{one real root}

\textsf{when }\:b^2-4ac < 0 \implies \textsf{no real roots}

As we need to determine the value of k that will give <u>one solution</u>, set the <u>discriminant to zero</u>.

<u>Given equation</u>:

y=kx^2-4x+4

Therefore:

  • a = k
  • b = -4
  • c = 4

Substitute these values into the discriminant and solve for k:

\begin{aligned}b^2-4ac & = 0\\\implies (-4)^2-4(k)(4) & = 0\\16-16k & = 0\\16k & = 16\\\implies k & = 1\end{aligned}

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2 years ago
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3 years ago
The lengths (in miles) of two highways after t years of construction can be modeled by 20t + 150 and 15t+ 200. What is the diffe
Mashutka [201]

Step-by-step explanation:

13.5 is the difference of the length of the roads

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2 years ago
The sum of 32 and what number is 105
natta225 [31]
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3 years ago
In the figure, TP and TS are opposite rays. TQ bisects &lt; 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

0

and ∠QRT=65

0

, then find the values of x and y.

463685

expand

Medium

Solution

verified

Verified by Toppr

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
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