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nikklg [1K]
3 years ago
15

How do you write 29.25 as an integer

Mathematics
1 answer:
Naily [24]3 years ago
6 0

Answer:

Step-by-step explanation:

lowercase:  twenty-nine and twenty-five hundredths

~ or, simpler ~  twenty-nine point twenty-five

~ or, even simpler ~

twenty-nine point two five

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Use the graph to find out over what interval f(x) is less than or equal to 0. PLEASE HELP ASAP!!!
jok3333 [9.3K]

Answer:

[-5, 2]

Step-by-step explanation:

We have to find the interval for which f(x) <= 0, the interval is [-5, 2].

5 0
3 years ago
Solve the equation by completing the square. Round to the nearest hundredth if necessary. x2 – x – 7 = 0
Tanzania [10]
2x-x-7=0

3x-7=0

add 7 to 7 and 0, so then we cross out 7-7, and then

3x=7, bc u change the subtraction to addition, sr if thst doesn't make sense but 3x=7 is the answer
8 0
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Sedbober [7]

Answer:

E ( 1 , -1 ) is in quadrant IV

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3 years ago
Solve the inequality and graph the solution<br> 7x ≥ -91
Alex Ar [27]

The solution is x \geq 13.

Solution:

Given inequality:

7 x \geq 91

Divide by 7 on both sides.

$\frac{7 x}{7} \geq \frac{91}{7}

x \geq 13

The solution is x \geq 13.

The image of the graph is attached below.

5 0
3 years ago
Select the two values of x that are roots of this equation 3x^2 + 1 =5x
Zina [86]

Answer:

x = 1.434 and x=0.232

Step-by-step explanation:

To find the root of the equation stated above we need to:

(1) Write the polynomial equation with zero on the right hand side:

3x^{2} + 1 = 5x ⇒ 3x^{2} -5x + 1 = 0

(2) Divide the whole equation by 3

3x^{2} -5x + 1 = 0 ⇒ x^{2} -\frac{5}{3}x + \frac{1}{3}= 0

(3) Use the quadratic formula to solve the quadratic equation:

The quadratic formula states that the two solutions for a quadratic equation is given by:

\frac{-b±\sqrt{b^{2} - 4ac}}{2a} (1)

In this case, a = 1, b = -\frac{5}{3}, c= \frac{1}{3}

Substituiting a, b and c in equation (1) We get:

\frac{-\frac{5}{3}±\sqrt{(-\frac{5}{3})^{2} - 4(1)(\frac{1}{3})}}{2(1)} (1)

The two solutions are:

x = 1.434 and x=0.232

8 0
3 years ago
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