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3 years ago

14. Andrew earns $6.25 per hour. He works 4 hours per day, 5 days

Mathematics

1 answer:

swat323 years ago

6 0

Answer:

$125 per week

Step-by-step explanation:

Andrew earns $6.25 per hour. If he works for 4 hours, solution will be:

$6.25 × 4 = $25 per day.

If Andrew works 5 days per week, then he earns

$25 × 5 = $125 per week.

Therefore by estimate, Andrew earns $125 per week.

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13. If Kelloggs wants to double all the dimensions of its Frosted Flakes box to create a "Costco" size box, how

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2 times as much.

(In the cereal aisle at Costco they always have 2 boxes a pack)

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3 years ago

Plz help im helping my brother :)

Tju [1.3M]

Answer:

Step-by-step explanation:

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Mr Barton was answering the problem below the question

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No the student is incorrect the answer is actually 332

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4 years ago

How would I solve 16x^4-41x^2+25=0 ???

sveticcg [70]

$16{ x }^{ 4 }-41{ x }^{ 2 }+25=0$

${ x }^{ 4 }={ ({ x }^{ 2 }) }^{ 2 }\\ \\ 16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0$

First of all to make our equation simpler, we'll equal $x^{2}$ to a variable like 'a'.

So,

${ x }^{ 2 }=a$

Now let's plug $x^{2}$ 's value (a) into the equation.

$16{ ({ x }^{ 2 }) }^{ 2 }-41{ x }^{ 2 }+25=0\\ \\ { x }^{ 2 }=a\\ \\ 16{ (a) }^{ 2 }-41{ a }+25=0$

Now we turned our equation into a quadratic equation.

(The variable 'a' will have a solution set of two solutions, but 'x' , which is the variable of our first equation will have a solution set of four solutions since it is a quartic equation (fourth-degree equation) )

Let's solve for a.

The formula used to solve quadratic equations ;

$\frac { -b\pm \sqrt { { b }^{ 2 }-4\cdot t\cdot c } }{ 2\cdot t }$

The formula is used in an equation formed like this :

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

In our equation,

t=16 , b=-41 and c=25

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

$t=16\quad b=-41\quad c=25\\ \\ \frac { -(-41)\pm \sqrt { -(41)^{ 2 }-4\cdot 16\cdot 25 } }{ 2\cdot 16 } \\ \\ \frac { 41\pm \sqrt { 1681-1600 } }{ 32 } \\ \\ \frac { 41\pm \sqrt { 81 } }{ 32 } \\ \\ \frac { 41\pm 9 }{ 32 }$

So the solution set :

$\frac { 41+9 }{ 32 } =\frac { 50 }{ 32 } \\ \\ \frac { 41-9 }{ 32 } =\frac { 32 }{ 32 } =1\\ \\ a\quad =\quad \left\{ \frac { 50 }{ 32 } ,\quad 1 \right\}$

We found a's value.

Remember,

${ x }^{ 2 }=a$

So after we found a's solution set, that means.

${ x }^{ 2 }=\frac { 50 }{ 32 }$

and

${ x }^{ 2 }=1$

We'll also solve this equations to find x's solution set :)

${ x }^{ 2 }=\frac { 50 }{ 32 } \\ \\ \frac { 50 }{ 32 } =\frac { 25 }{ 16 } \\ \\ { x }^{ 2 }=\frac { 25 }{ 16 } \\ \\ \sqrt { { x }^{ 2 } } =\sqrt { \frac { 25 }{ 16 } } \\ \\ x=\quad \pm \frac { 5 }{ 4 }$

${ x }^{ 2 }=1\\ \\ \sqrt { { x }^{ 2 } } =\sqrt { 1 } \\ \\ x=\quad \pm 1$

So the values x has are :

$\frac { 5 }{ 4 }$ , $-\frac { 5 }{ 4 }$ , $1$ and $-1$

Solution set :

$x=\quad \left\{ \frac { 5 }{ 4 } \quad ,\quad -\frac { 5 }{ 4 } \quad ,\quad 1\quad ,\quad -1 \right\}$

I hope this was clear enough. If not please ask :)

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Find the area of the following figure. Explain how you got your answer.

steposvetlana [31]

Answer:

26.5 units. You count all the units and then count the halves

Step-by-step explanation:

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