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dsp73
2 years ago
5

How would i solve this? SOLVE FOR TU

Mathematics
1 answer:
creativ13 [48]2 years ago
7 0
First subtract 30-23 that’d be the value of UV(7) now subtract 18 from 7, the answer will be 11 therefore TU is 11.
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Why is it undefined for the equation 0/0?
kodGreya [7K]

Answer:

Because anything divided by 0 is undefined. For example, 7/0 is undefined because 7 is being divided by 0 and that cant happen.

8 0
3 years ago
Read 2 more answers
Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat
zavuch27 [327]

Answer:

The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term x^2, which means that its graph shows branches going up. Therefore:

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

2) if its vertex is below the x-axis, it is 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 not have 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=1 and b=5, we get that the x-position of the vertex is: x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the x_v=-\frac{5}{2}. Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.

When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):

f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0

The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.

Then, the two roots are: x = -1 and x = -4.

5 0
3 years ago
What is the area of that shaded sector?
melomori [17]

Answer:

50.24

Step-by-step explanation:

1/4x3.14xr^2

r=16 because of the length

90/360= 1/4

so

1/4x3.14x16^2/4

=50.24

4 is our denominator because of the denominator on the fraction

4 0
3 years ago
I really need help quick I’ve been working this out for 1 hour already
wolverine [178]

Answer:

B

Step-by-step explanation:

Hope this helps

4 0
3 years ago
Analyze the graph h(x).
Dominik [7]

Answer:

D h(x) = f(x)×g(x)

Step-by-step explanation:

h(x) has a wave with 2 changes in direction.

so, this needs to be an expression of the third degree (there must be a term with x³ as the highest power of x).

and that is only possible when multiplying both basic functions. all the other options would keep it at second degree (x²) or render it even to a first degree (linear).

8 0
3 years ago
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