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4 years ago

Emilio’s paycheck was $305.25, and he makes $9.25 per hour. How many hours did Emilio work?

Mathematics

2 answers:

GuDViN [60]4 years ago

6 0

305.25/9.25=33 hours

Anastaziya [24]4 years ago

5 0

$305.25 divided by $9.25 equals 33, so he worked 33 hours... hope this helped.

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Sphinxa [80]

The answer would be B

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2 years ago

What are the real solutions of x squared =225

Gnesinka [82]

The real solutions the equation as given in the task content; x² = 225 are; +25 and -25.

<h3>What are the real solutions of the equation as given in the task content?</h3>

It follows from the task content that the real solutions of the equation as given in the task content can be determined as follows;

x² = 225

x = ± 15

Therefore, the real solutions of the equation are; +25 and -25.

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2 years ago

HELP IM ON A TIME LIMIT AND IDK THE ANSWER

aleksandrvk [35]

Answer:

Check step-by-step explanation

Step-by-step explanation:

A = $\frac{1}{2}$ $d_{1} d_{2}$

to solve for $d_{1}$ :

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$d_{2}$ = $\frac{2A}{d_{1}}$

4 0

3 years ago

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Gnoma [55]

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3 years ago

What are the possible rational roots of the polynomial equation?<br><br> 0=2x7+3x5−9x2+6

RoseWind [281]

Answer: $\pm\frac{1}{1}, \pm\frac{1}{2},\pm\frac{2}{1},\pm\frac{3}{1}, \pm\frac{3}{2}$

Step-by-step explanation:

We can use the Rational Root Test.

Given a polynomial in the form:

$a_nx^n +a_{n- 1}x^{n - 1} + … + a_1x^1 + a_0 = 0$

Where:

- The coefficients are integers.

- $a_n$ is the leading coeffcient ( $a_n\neq 0$ )

- $a_0$ is the constant term $a_0\neq 0$

Every rational root of the polynomial is in the form:

$\frac{p}{q}=\frac{\pm(factors\ of\ a_0)}{\pm(factors\ of\ a_n)}$

For the case of the given polynomial:

$2x^7+3x^5-9x^2+6=0$

We can observe that:

- Its constant term is 6, with factors 1, 2 and 3.

- Its leading coefficient is 2, with factors 1 and 2.

Then, by Rational Roots Test we get the possible rational roots of this polynomial:

$\frac{p}{q}=\frac{\pm(1,2,3,6)}{\pm(1,2)}=\pm\frac{1}{1}, \pm\frac{1}{2},\pm\frac{2}{1},\pm\frac{3}{1}, \pm\frac{3}{2}$

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3 years ago