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Advocard [28]
3 years ago
6

Divide.

Mathematics
2 answers:
vovikov84 [41]3 years ago
5 0
Before you can divide anything mixed numbers, you must first put it as an improper fraction so 5 1/4 will change into 21/4 because 5*4+1 is 21 and don’t forget the fraction so it would be 21/4, and -2 1/2 would change into -5/4 because -2*2+1 is -5 and if u add the fraction it would become -5/4. Now, the method I use is leave, change, flip, so I first leave the 21/4 in 21/4/-5/4 and I change the division sign into a multiplication sign, and I flip the -5/4 to -4/5, so your new equation now looks like, 21/4•-4/5. Now we have to multiply. Since we can cross simplify, you can cancel out both fours leaving you with 21/1•-1/5 and when u multiply across you get -21/5. When u simplify it to a mixed number, it should look like -4 1/5 because you see how many times 5 goes into 21 which is 4 times and it leaves u with a remainder of 1 because 21-20 is 1 and when ever u have a remainder, u have to keep a denominator of 5 so your final answer is: -4 1/5
patriot [66]3 years ago
3 0

Answer:

The correct answer is -2 1/10

^-^


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What number does x stand for in this equation?
Alex777 [14]

Answer:

C. -3

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

Step-by-step explanation:

<u>Step 1: Define</u>

-7x + 6 = 27

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. [Subtraction Property of Equality] Subtract 6 on both sides:                        -7x = 21
  2. [Division Property of Equality] Divide -7 on both sides:                                 x = -3
3 0
2 years ago
Y
8090 [49]

The area of the triangle will be 24912 sq. units. Square units and other similar units are used to measure area.

<h3>What is the area?</h3>

The space filled by a flat form or the surface of an item is known as the area.

The number of unit squares that cover the surface of a closed-form is the figure's area.

For:

(X1, Y1) = (1, 15)

(X2, Y2) = (-2, 1)

d = 14.317821

For:

(X₂, Y₂) = (-2, 1)

(X₃, Y₃) = (4, 5)

d = 7.211103

For applying the pythogorous them we need the right angle triangle obtained by bisect from the mid point.

The value of the base is;

⇒7.2 / 2

⇒3.6

apply the pythogorous theorem for finding the height;

h² = p² + b²

14.31² = p² + 3.6²

p = 13.84

The area of the triangle is;

\rm A = \frac{1}{2}\times b \times h \\\\ A= \frac{1}{2} \times 3.6 \times 13.84 \\\\ A = 24.912

Hence, the area of the triangle will be 24912 sq. units.

To learn more about the area, refer to the link;

brainly.com/question/11952845

#SPJ1

6 0
1 year ago
How exactly do you do this because my teacher didn't teach me how
mylen [45]
The area is found by base x height

To find the missing height in the first one you divide 147 by 15 3/4 and get 9 1/3. 
To find the missing base in the second one you divide 140 5/8 by 11 1/4 and get 12 1/2.
To find the missing height in the third one you divide 151 3/16 by 10 1/4 and get 14 3/4. 

Make me brainliest ! :)

6 0
3 years ago
PLEASE HELP PLESD PLED​
-BARSIC- [3]

Answer:

<h2>x = 35</h2>

Step-by-step explanation:

Because the angles are vertical angles, they must be equal.

3x + 7 = 112

3x = 105

x = 35

Check

3(35) + 7 = 112

112 = 112 Correct

7 0
2 years ago
Read 2 more answers
D<br> Evaluate<br> arcsin<br> (6)]<br> at x = 4.<br> dx
sineoko [7]

Answer:

\frac{1}{2\sqrt{5} }

Step-by-step explanation:

Let, \text{sin}^{-1}(\frac{x}{6}) = y

sin(y) = \frac{x}{6}

\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})

\frac{d}{dx}\text{sin(y)}=\frac{1}{6}

\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx} ---------(1)

\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}

\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}

cos(y) = \sqrt{1-\text{sin}^{2}(y) }

          = \sqrt{1-(\frac{x}{6})^2}

          = \sqrt{1-(\frac{x^2}{36})}

Therefore, from equation (1),

\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

Or \frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

At x = 4,

\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}

\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}

                   =\frac{1}{6\sqrt{\frac{36-16}{36}}}

                   =\frac{1}{6\sqrt{\frac{20}{36} }}

                   =\frac{1}{\sqrt{20}}

                   =\frac{1}{2\sqrt{5}}

4 0
2 years ago
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