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zimovet [89]
3 years ago
12

(2x² + 3x - 5)(-x² + 7x + 2)​

Mathematics
1 answer:
Alexxandr [17]3 years ago
6 0

Answer:

-2x^4 + 11x^3 + 30x^2 - 29x - 10

Step-by-step explanation:

1. Use the Distributive Property

(2x^2 + 3x - 5)(-x^2 + 7x + 2)

2. Remove parenthesis

2x^2(-x^2 + 7x + 2) + 3x(-x^7x+2) - 5(-x^2 + 7x + 2)

3. Collect like terms

-2x^4 + 14x^3 + 4x^2 - 3x^3 + 21x^2 + 6x + 5x^2 - 35x - 10

4. -2x^4 + 11x^3 + 30x^2 - 29x - 10

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What is the greatest number of rìght angles that a triangle can contain?
sergiy2304 [10]

Answer:

1

Step-by-step explanation:

If there are 2 or more right angles in a triangle, it is impossible to connect without a fourth side, making it a square. There can only be 1 right angle in a triangle

5 0
3 years ago
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Please help will give 20 points to right answers (will give brainliest to very first correct answer) Tom walks 3.2 km every morn
nignag [31]
For this case we have the following conversions:
 1 mile = 1,609 km

1 mile = 5,280 feet
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6 0
3 years ago
Could someone help me real quick.
Keith_Richards [23]
12 is the correct answer. When you have problems like this just add over the equal sign. So 5 plus 6 plus 1 is 12. 12 minus 6 minus 1 is 5!
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5 0
3 years ago
The coordinates G(7,3), H(9, 0), (5, -1) form what type of polygon?
Marrrta [24]

Answer:

Is an acute triangle

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

we have

G(7,3), H(9, 0), I(5, -1)

step 1

Find the distance GH

substitute in the formula

d=\sqrt{(0-3)^{2}+(9-7)^{2}}

d=\sqrt{(-3)^{2}+(2)^{2}}

GH=\sqrt{13}\ units

step 2

Find the distance IH

substitute in the formula

d=\sqrt{(0+1)^{2}+(9-5)^{2}}

d=\sqrt{(1)^{2}+(4)^{2}}

IH=\sqrt{17}\ units

step 3

Find the distance GI

substitute in the formula

d=\sqrt{(-1-3)^{2}+(5-7)^{2}}

d=\sqrt{(-4)^{2}+(-2)^{2}}

GI=\sqrt{20}\ units

step 4

Verify what type of triangle is the polygon

we know that

If applying the Pythagoras Theorem

c^{2}=a^{2}+b^{2} ----> is a right triangle

c^{2}> a^{2}+b^{2} ----> is an obtuse triangle

c^{2}< a^{2}+b^{2} ----> is an acute triangle

where

c is the greater side

we have

c=\sqrt{20}\ units

a=\sqrt{17}\ units

b=\sqrt{13}\ units

substitute

c^{2}= (\sqrt{20})^{2}=20

a^{2}+b^{2}=(\sqrt{17})^{2}+(\sqrt{13})^{2}=30

therefore

c^{2}< a^{2}+b^{2}

Is an acute triangle

6 0
3 years ago
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