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2 years ago

Oliver earns $23.80 an hour. He earns normal rates during week days and time-and-a-

1 answer:

8 0

So first we calculate the weekly hours:
23.80 x 34 = 809.20

Then we calculate the time and half:
23.80 x 6 x 1.5 = 214.2

Then we total then together:
214.2 + 809.20 = 1023.40

You might be interested in

Find the square root of 13²

nadezda [96]

The answer to that would be “169”

3 0

3 years ago

Please help! Thanks in advance.

alukav5142 [94]

In this problem, we need to plug in the given x values for $f(x)=0$ and find a and b.

When we plug in 1, we get: $2 \times {1}^{4} - 5 \times {1}^{3} - 14 \times {1}^{2} + a \times 1 + b = 0$

Simplify:
$2 - 5 - 14 + a + b = 0$
$- 17 + a + b = 0$
$a + b = 17$

We got our first statement about the values of the variables. If we find one more we can find those 2 variables.

We have another given root: 4.

Plug it in:
$2 \times {4}^{4} - 5 \times {4}^{3} - 14 \times {4}^{2} + a \times 4 + b = 0$
$512 - 320 - 224 + 4a + b = 0$
$- 32 + 4a + b = 0$
$4a + b = 32$

Now we have our second one. We can combine them:
$a + b = 17 \\ 4a + b = 32$

I use elimination method which is easier here.

Multiply the top equation by -1: $- a - b = - 17 \\ 4a + b = 32$

Add them up:
$3a = 15$

Simplify:
$a = 5$

Now we have a, we can plug in one of those equations to find b:
$5 + b = 17$
$b = 17 - 5$
$b = 12$

So, the answers are $a=5$ and $b=12$ .

6 0

3 years ago

Can i have an answer

nordsb [41]

Answer:

I think it would be b

Step-by-step explanation:

im sorry if im wrong im not good with this stuff :c

5 0

3 years ago

Need help asap thanks

Ymorist [56]

Answer:

Step-by-step explanation:

for a right angled triangle, the sum of the square of the two sides should be equal to the square of the hypotenuse. If this rule does not hold, the dimensions do not represent the side of a right angled triangle.

The Pythagoras theorem : a² + b² = c²

where a = length

b = base

c = hypotenuse

4² + 10² ≠ 16²

6² + 12² = 13²

7² + 24² = 25²

8² + 15² = 17²

8 0

2 years ago

Can you help me please??

dusya [7]