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3 years ago

6(5r-11)-(5-r)}6(5r−11)−(5−r)

Mathematics

2 answers:

Juli2301 [7.4K]3 years ago

5 0

30r(2)-185r+259=3r(2)-185r+259

denpristay [2]3 years ago

3 0

Answer:

31r-71

Step-by-step explanation:

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6. Find the area of the triangle below.<br> Be sure to include the correct unit in your answer.

snow_tiger [21]

Answer:

512

Step-by-step explanation: area of triangle= 1/2*base*height

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3 years ago

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Can someone help me with question e please ?

MAXImum [283]

Answer:

Step-by-step explanation:

a) The domain is R- all real numbers.

b) The domain is R- all real numbers.

c) we need

$x^{2}-4\neq 0\\x^{2} \neq 4\\x\neq 2\\and\\x\neq -2$

So the domain is R\{-2, 2}

we need

$x^{2}+x-2\neq 0\\x\neq 1\\and\\x\neq -2$

So the domain is R\{-2, 1}

we need: x-1>0 or x> 1

So the domain is (1,+∞)

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3 years ago

Macie Andersen earns $9.75 an hour working at the local bookstore. Last week she worked 17 hours. This week she worked 25 hours.

vovangra [49]

Answer:

$407.50

Step-by-step explanation:

take 9.75 × 17=165.75

take 9.75 × 25=243.75

add 165.75 and 243.75=407.50

turn into dollars $407.50

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3 years ago

. You draw a card from a well-shuffled deck of 52 cards. If you get a red card, you win nothing. If you get a spade, you win $4.

Novosadov [1.4K]

Answer is in the file below

tinyurl.com/wtjfavyw

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3 years ago

Triangle ABC has vertices at A(−1 , 2), B(4, 2), and C(3, −1). Classify the triangle according to the side lengths.

Orlov [11]

Answer:

B) Isosceles

Step-by-step explanation:

The given triangle has vertices at A(-1,2), B(4,2) and C(3,-1).

We must first determine the length of the sides of the triangle, before we can classify it.

We apply the distance formula to find length of the sides.

$|AB|=\sqrt{(4--1)^2+(2-2)^2}$

$\Rightarrow |AB|=\sqrt{(4+1)^2+(2-2)^2}$

$\Rightarrow |AB|=\sqrt{5^2+(0)^2}$

$\Rightarrow |AB|=\sqrt{25}$

$\Rightarrow |AB|=5 units$ .

The length of side BC

$|BC|=\sqrt{(3-4)^2+(-1-2)^2}$

$\Rightarrow |BC|=\sqrt{(-1)^2+(-3)^2}$

$\Rightarrow |BC|=\sqrt{1+9}$

$\Rightarrow |BC|=\sqrt{10}$

The length of side AC

$|AC|=\sqrt{(3--1)^2+(-1-2)^2}$

We simplify to obtain;

$|AC|=\sqrt{(3+1)^2+(-3)^2}$

$\Rightarrow |AC|=\sqrt{(4)^2+(-3)^2}$

$|AC|=\sqrt{16+9}$

$|AC|=\sqrt{25}$

$|AC|=5\:units$

Since $|AC|=5\:units=|AB|$ , the given triangle is an isosceles triangle.

The correct answer is

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3 years ago

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