Select the correct answer.

Round 584,565 to the nearest hundred thousand.

lora16 [44]

Answer:

600,000

Step-by-step explanation:

600,000

8 0

3 years ago

Read 2 more answers

-x+4=x+6 i have to solve for x

Andreas93 [3]

Answer:

Collect like terms

-x-x=6-4

Solve

-2x=2, -x=1 or x=-1

Step-by-step explanation:

I hope this helps, please give me brainliest

6 0

3 years ago

The force of gravity on an object varies directly with its mass. The constant of variation due to gravity is 32.2 feet per secon

castortr0y [4]

Answer:

Step-by-step explanation:

According to newton second law, the force of gravity on an object varies directly with its mass and it is expressed mathematically as Fαm i.e

F = mg where;

F is the force of gravity

m is the mass of the body

g is the proportionality constant known as the acceleration due to gravity.

If the constant of variation due to gravity is 32.2ft/s², the equation that represents F, the force on an object due to gravity according to m, the object’s mass can be gotten by substituting g = 32.2 into the formula above according to the law as shown;

F = m*32.2

F =32.2m

Hence the required equation is F = 32.2m

5 0

3 years ago

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A farmer grows vegetables on a rectangular plot of land. The length of the plot is labeled in the diagram below. If the perimete

gavmur [86]

Answer:

Step-by-step explanation:

solve the question then divide

4 0

3 years ago

The function f(x) = 2x2 + 3x + 5, when evaluated, gives a value of 19. What is the function’s input value?

goldenfox [79]

$f(x)=\quad 2{ x }^{ 2 }+3x+5$

Let's called the input 'z'

When we plug 'z' in the function we get ;

$f(z)=\quad 2{ z }^{ 2 }+3z+5$

And we know that, this is equal to 19, so ;

$2{ z }^{ 2 }+3z+5=\quad 19$

Let's rearrange the equation.

$2{ z }^{ 2 }+3z+5=\quad 19\\ \\ 2{ z }^{ 2 }+3z=\quad 19-5\\ \\ 2{ z }^{ 2 }+3z=\quad 14\\ \\ 2{ z }^{ 2 }+3z-14=\quad 0$

So we have a quadratic equation here.

We'll use this formula to solve it :

$\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

The formula is used in equation formed like this :

$a{ x }^{ 2 }+bx+c=0$

In our equation,

a=2 , b=3 and c=-14

Let's plug in the values in the formula to solve,

$a=2\quad b=3\quad c=-14\\ \\ \frac { -3\pm \sqrt { 9-(4\cdot 2\cdot -14) } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 9-(-112) } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 9+112 } }{ 4 } \\ \\ \frac { -3\pm \sqrt { 121 } }{ 4 } \\ \\ \frac { -3\pm 11 }{ 4 }$

So,

$z=\quad \frac { -3+11 }{ 4 } \quad ,\quad \frac { -3-11 }{ 4 } \\ \\ z=\quad \frac { 8 }{ 4 } \quad ,\quad \frac { -14 }{ 4 } \\ \\ z=\quad 2,\quad -\frac { 7 }{ 2 }$

So the input can be both, 2 and $-\frac{7}{2}$

8 0

4 years ago