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2 years ago

1. Is triangle P’Q’R larger or smaller than triangle PQR? Explain how you know

Mathematics

1 answer:

Salsk061 [2.6K]2 years ago

7 0

Answer:

The length of the line is PQ as this line is parallel to the x - axis. So, the length of the line is the summation of 10 from second quadrant and 20 from first quadrant. So, the sum is 30. Hence the length of the line is 30 units.

Step-by-step explanation:

The length of a line segment can be measured by measuring the distance between its two endpoints. It is the path between the two points with a definite length that can be measured. Explanation: On a graph, the length of a line segment can be found by using the distance formula between its endpoints.

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Use the quadratic formula to find the solutions for.. y = 2x^2 - 9x + 5.

Yakvenalex [24]

Answer: $x=\frac{9}{4}-\frac{\sqrt{41} }{4} \\x=\frac{9}{4}+\frac{\sqrt{41} }{4}$

Step-by-step explanation:

$2x^2-9x+5$

a=2

b=-9

c=5

$x=\frac{-b\frac{+}{}\sqrt{b^2-4ac} }{2a}$

$x=\frac{-(-9)\frac{+}{}\sqrt{(-9)^2-4(2)(5)} }{2(2)}$

$x=\frac{9\frac{+}{}\sqrt{81-40} }{4}$

$x=\frac{9\frac{+}{}\sqrt{41} }{4}$

$x=\frac{9}{4}-\frac{\sqrt{41} }{4} \\x=\frac{9}{4}+\frac{\sqrt{41} }{4}$

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2 years ago

What is the quotient in simplest form?

enyata [817]

The third one 02 2/5 is the answer

8 0

2 years ago

What is the measure of ∠DEG? A) 133° B) 103° C) 77° D) 67°

nordsb [41]

Hmmmmmm
Let me think for a second

3 0

3 years ago

Read 2 more answers

What are the solutions to the equation

frosja888 [35]

Answer:

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

Step-by-step explanation:

You have the quadratic function $2x^2-x+1=0$ to find the solutions for this equation we are going to use Bhaskara's Formula.

For the quadratic functions $ax^2+bx+c=0$ with $a\neq 0$ the Bhaskara's Formula is:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}$

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}$

It usually has two solutions.

Then we have $2x^2-x+1=0$ where a=2, b=-1 and c=1. Applying the formula:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}$

Observation: $\sqrt{-1}=i$

$x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$

And,

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i$

Then the correct answer is option C.

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

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2 years ago

(11w^4 × 3z^9)(2w^7z^2)

mart [117]

Simplified:

11w^4(3)z^9(2w^7z^2)

=66w^11z^11

4 0

2 years ago