Evaluate (7.1x 10³)+(4.6 x 10⁶)² <br>Give your answer in standard form to 3SF

What kind of solution is -4x-4=-7(x+4) ?

natima [27]

~ Simplifying

-4x + -4 = -7(x + 4)

~ Reorder the terms:

-4 + -4x = -7(x + 4)

~ Reorder the terms:

-4 + -4x = -7(4 + x)

-4 + -4x = (4 * -7 + x * -7)

-4 + -4x = (-28 + -7x)

~ Solving

-4 + -4x = -28 + -7x

~ Solving for variable 'x'.

~ Move all terms containing x to the left, all other terms to the right.

~ Add '7x' to each side of the equation.

-4 + -4x + 7x = -28 + -7x + 7x

~ Combine like terms: -4x + 7x = 3x

-4 + 3x = -28 + -7x + 7x

~ Combine like terms: -7x + 7x = 0

-4 + 3x = -28 + 0

-4 + 3x = -28

~ Add '4' to each side of the equation.

-4 + 4 + 3x = -28 + 4

~ Combine like terms: -4 + 4 = 0

0 + 3x = -28 + 4

3x = -28 + 4

~ Combine like terms: -28 + 4 = -24

3x = -24

~ Divide each side by '3'.

x = -8

~ Simplifying

x = -8

8 0

3 years ago

Read 2 more answers

Help QUICKLY, will offer brainliest for correct answer

FromTheMoon [43]

Answer:

48

Step-by-step explanation:

-2(1)^2(3(1)^2 – 7(1) + 10)

-2^2(3^2 – 7 + 10)

4(9 - 7 + 10)

4(12)

48

4 0

3 years ago

Olivia rides her scooter 3/4 mile in 1/3 hour. How fast, in miles per hour, does she ride her scooter?

Aleonysh [2.5K]

33/4 = 8-1/4 miles/ hour

7 0

2 years ago

Read 2 more answers

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation $(x+0,y-2)$ will map vertices of the square into points

$(2,1)\rightarrow (2,-1);$
$(1,2)\rightarrow (1,0);$
$(2,3)\rightarrow (2,1);$
$(3,2)\rightarrow (3,0).$

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

$(2,-1)\rightarrow (2,3);$
$(1,0)\rightarrow (1,2);$
$(2,1)\rightarrow (2,1);$
$(3,0)\rightarrow (3,2)$

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation $(x+2,y-0)$ will map vertices of the square into points

$(2,1)\rightarrow (4,1);$
$(1,2)\rightarrow (3,2);$
$(2,3)\rightarrow (4,3);$
$(3,2)\rightarrow (5,2).$

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

$(4,1)\rightarrow (2,1);$
$(3,2)\rightarrow (3,2);$
$(4,3)\rightarrow (2,3);$
$(5,2)\rightarrow (1,2)$

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation $(x+3,y+3)$ will map vertices of the square into points

$(2,1)\rightarrow (5,4);$
$(1,2)\rightarrow (4,5);$
$(2,3)\rightarrow (5,6);$
$(3,2)\rightarrow (6,5).$

2nd transformation = reflection over y = -x + 7 will map vertices into points

$(5,4)\rightarrow (3,2);$
$(4,5)\rightarrow (2,3);$
$(5,6)\rightarrow (1,2);$
$(6,5)\rightarrow (2,1)$

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation $(x-3,y-3)$ will map vertices of the square into points

$(2,1)\rightarrow (-1,-2);$
$(1,2)\rightarrow (-2,-1);$
$(2,3)\rightarrow (-1,0);$
$(3,2)\rightarrow (0,-1).$

2nd transformation = reflection over y = -x + 2 will map vertices into points

$(-1,-2)\rightarrow (4,3);$
$(-2,-1)\rightarrow (3,4);$
$(-1,0)\rightarrow (2,3);$
$(0,-1)\rightarrow (3,2)$

These points are not the vertices of the initial square.

5 0

3 years ago

#4 NEAD HELP SOLVE THIS QUESTION ASAP

sergiy2304 [10]

C. 118.2

1 and 2 make a complimentary angle, so 2=118.2

2 and 6 are corresponding, so angle 2=angle 6 (118.2 = 118.2)

4 0

3 years ago

Evaluate (7.1x 10³)+(4.6 x 10⁶)² Give your answer in standard form to 3SF​

Evaluate (7.1x 10³)+(4.6 x 10⁶)² Give your answer in standard form to 3SF