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melisa1 [442]
3 years ago
12

Deepak wrote out the steps to his solution of the equation StartFraction 5 Over 2 minus 3 x minus 5 plus 4 x equals negative Sta

rtFraction 7 Over 4 EndFraction – 3x – 5 + 4x = –. Which step has an incorrect instruction? Step 1 Step 2 Step 3 Step 4
Mathematics
2 answers:
Anna11 [10]3 years ago
8 0

Answer:

step1

Step-by-step explanation: yes   sirr

Anna35 [415]3 years ago
6 0

Answer:

Step 1 is incorrect.

Step-by-step explanation:

The given expression is :

\dfrac{5}{2}-3x-5+4x=\dfrac{-7}{4}

In step 1, Deepak rearranged the terms in order to simplify like terms together that is followed in step 2.

o, rearranging of the terms of an expression is commutative property and not distributive property.  The distributive property is generally applicable to multiplication and is given as:

x(y+z)=xy+yz

The commutative property is given as:

x+y = y +a

Therefore, adding terms by changing their order is commutative property of addition and not distributive property. Step 1 instruction is wrong.

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What is 7,230,000,000 in scientific notation <br> What is 0.0000865 in scientific notation
Gekata [30.6K]

Answer:

7,230,000,000 = 7.23 x 10 to the 9th power. 0.0000865 = 8.65 x 10 to the negative 5th power.

Step-by-step explanation:

7 0
3 years ago
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Insert a digit to make the number divisible by 24 if possible:<br> 44_8<br> pls help
Nat2105 [25]

Answer:

it's 4488 Wich would make 187

5 0
3 years ago
A.
Savatey [412]

Answer:

E

Step-by-step explanation:

(X1, Y1) = (-3,-2)

(X2,Y2)=(4,8)

SLOPE= (Y2-Y1) / (X2-X1)

= (8-(-2)) / (4-(-3))

=(8+2) / (4+3)

= 10/7

Hope it was helpful

5 0
3 years ago
In ΔJKL, j = 74 cm, k = 14 cm and l=80 cm. Find the measure of ∠J to the nearest degree.
natita [175]

Answer:

  ∠J = 60°

Step-by-step explanation:

The Law of Cosines tells you ...

  j² = k² +l² -2kl·cos(J)

Solving for J gives ...

  J = arccos((k² +l² -j²)/(2kl))

  J = arccos((14² +80² -74²)/(2·14·80)) = arccos(1120/2240) = arccos(1/2)

  J = 60°

_____

<em>Additional comment</em>

It is pretty rare to find a set of integer side lengths that result in one of the angles of the triangle being a rational number of degrees.

6 0
2 years ago
Quadrilateral EFGH has coordinates E(2a, 2a), F(3a, a), G(2a, 0), and H(0, 0). Find the midpoint of he
Gre4nikov [31]
The formula of the midpoint of HE:
\left(\dfrac{x_H+x_E}{2};\ \dfrac{y_H+y_E}{2}\right)

We have H(0; 0) and E(2a; 2a). Substitute:
\left(\dfrac{0+2a}{2};\ \dfrac{0+2a}{2}\right)=(a;\ a)

4 0
3 years ago
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