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Alexxandr [17]
3 years ago
7

To the nearest hundreth What is the relative

Mathematics
1 answer:
Alja [10]3 years ago
6 0

Answer:

i think the right answer is the first one

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Tomorrow the last day of school how ya feel?<br> -franz:)
zheka24 [161]

Answer:

Ours is today. Feels weird how we came a long way with the pandemic presiding our academic studies.

Step-by-step explanation:

6 0
3 years ago
Help me answer this.. plz
Fudgin [204]

Answer:

2\frac{1}{30}

Step-by-step explanation:

First thing you gotta do is find the common denominator.

Do this by listing factors of the denominators:

5- 5, 10, 15, 20, 25, 30

6- 6, 12, 18, 24, 30, 36

5 and 6 both have 30 in common so that's the new denominator for all 3 of the terms.

What you do to the bottom you must do to the top.

Since you multiply 5 by 6 to get 30, you must multiply 2 and 4 by 6 as well.

Since you multiply 6 by 5 to get 30, then multiply 5 by 5 as well.

The new equation is:

\frac{12}{30} +\frac{25}{30} + \frac{24}{30} =

Now add the numerators together

The solution is \frac{61}{30}. Now simplify. 61 is prime, so you cannot change the fraction How many times does 30 go into 61? Two times. This gives us 2\frac{1}{30}.

4 0
3 years ago
Find the slope from the following table: x 2 3 4 y 4 5 6
Sindrei [870]

Answer:

the slope is 1

Step-by-step explanation:

The linear equation is y=x+2

7 0
3 years ago
Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat
eimsori [14]

Answer:

There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.

Step-by-step explanation:

We notice that the expression on the left of the equation is a quadratic with leading term 2x^2, which means that its graph is that of a parabola with branches going up.

Therefore, there can be three different situations:

1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.

2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.

We recall that the x-position of the vertex for a quadratic function of the form  f(x)=ax^2+bx+c is given by the expression:

x_v=\frac{-b}{2a}

Since in our case a=2 and b=-3, we get that the x-position of the vertex is:

x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:

y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}

This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.

We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:

f(-1) = 2(-1)^2-3(-1)+4= 2+3+4=9\\f(0)=2(0)^2-3(0)+4=0+0+4=4\\f(1)=2(1)^2-3(-1)+4=2-3+4=3\\f(2)=2(2)^2-3(2)+4=8-6+4=6

See the graph produced in the attached image.

4 0
3 years ago
Find a complex number a+ ????????????????, with aand ???????? positive real numbers, such that (a+ ????????????????)3 = i.
stepan [7]

Answer:

The complex number is 0+i/3.

Step-by-step explanation:

Let the imaginary part be x.

The complex number will be a+xi

Given:

real part =a

(a+xi)*3=i...............(i)

To find:

imaginary part

Solution:

3a + 3xi=i.............(ii)

Comparing real and imaginary parts of eq(ii).

3a=0  and 3x=1

a=0

x=1/3.

6 0
3 years ago
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