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3 years ago

What is (7s^2+3s)-(4s^2-3s+1) in simplest form?

Mathematics

1 answer:

DENIUS [597]3 years ago

5 0

First distribute the negative over the parentheses:-
= 7s^2 + 3s - 4s^2 + 3s - 1
= 3s^2 +6s - 1

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Three U.F.O.'s were constructed in consecutive years. Their ages have a sum of

Dahasolnce [82]

Answer: 41, 42, and 43 years old.

Step-by-step explanation:

41 plus 42 plus 43 = 126.

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3 years ago

In a set of five consecutive integers, the smallest integer is more than $\frac23$ the largest. What is the smallest possible va

Goryan [66]

Answer:

Step-by-step explanation:

Let x represent the middle integer. Then the smallest is x-2 and the largest is x+2. Your requirement is that ...

(x-2)/(x+2) > 2/3

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The smallest sum is 55.

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3 years ago

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Solve the equation 8d – 4d – 6d – 8 = 2d A. 0 B. -4 C. -2 D. -4

skad [1K]

<span>8d – 4d – 6d – 8 = 2d

Combine common terms:
-2d - 8 = 2d

Add 2d to both sides:
-8 = 4d

Swtich sides:
4d = -8

Divide both sides by 4:
d = -2

Answer: - 2
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2 years ago

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Hello! Can someone please help me with these 2 questions! thank you so much! (i’ll give brainlist)

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Answer:

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Step-by-step explanation:

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2 years ago

Determine if triangle ABC with coordinates A (0, 2), B (2, 5), and C (−1, 7) is an isosceles triangle. Use evidence to support y

kirill115 [55]

Answer:

The triangle ABC is an isosceles right triangle

Step-by-step explanation:

we have

The coordinates of triangle ABC are

A (0, 2), B (2, 5), and C (−1, 7)

we know that

An isosceles triangle has two equal sides and two equal internal angles

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

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

step 1

Find the distance AB

substitute in the formula

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

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

$dAB=\sqrt{13}\ units$

step 2

Find the distance BC

substitute in the formula

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

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

$dBC=\sqrt{13}\ units$

step 3

Find the distance AC

substitute in the formula

$d=\sqrt{(7-2)^{2}+(-1-0)^{2}}$

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

$dAC=\sqrt{26}\ units$

step 4

Compare the length sides

$dAB=\sqrt{13}\ units$

$dBC=\sqrt{13}\ units$

$dAC=\sqrt{26}\ units$

$dAB=dBC$

therefore

Is an isosceles triangle

Applying the Pythagoras Theorem

$(AC)^{2} =(AB)^{2}+(BC)^{2}$

substitute

$(\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}$

$26=13+13$

$26=26$ -----> is true

therefore

Is an isosceles right triangle

8 0

3 years ago