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finlep [7]
3 years ago
13

Anyone know the answer???

Mathematics
1 answer:
alisha [4.7K]3 years ago
5 0

Hello from MrBillDoesMath!

Answer:

k = 10

Discussion:

angle MYK = 180 and angle MYX = 180 =>  angle ZYX = 180 - 115 - 65

Trialing XYZ has 180 degrees so

180 =  (4k + 5) +  (6k+10) + 65     => combine like terms

180 = (4k + 6k) + (5 + 10 + 65)     => as 4k +6k = 10k

180 = 10k  + 80                              => subtract 80 from both sides

180 -80 = 100 = 10k                      => divide both sides by 10

100/10 = k

Thank you,

MrB

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Answer:

The answer is B. The first number in the sequence is going to be 2 and the second is going to be 7. you can plug those numbers in the check your answer but I already did it lol. Hope this helps!!

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A package of twelve cans of the black beans cost $6.84. What is the cost of three cans of black beans?
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$1.66

Step-by-step explanation:

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3 years ago
Z-m=z+bx , solve for b
KIM [24]

Answer:

b= (Z-m-z)/(x)

Step-by-step explanation:

Z-m=z+bx

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3 0
3 years ago
Read 2 more answers
M Find the measure of each angle.
Illusion [34]

Answer:

Step-by-step explanation:

According the Triangle sum theorem:

m<T + m<U + m<V =180

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Therefore m<T = 25, m<U = 50, and m<V = 105

4 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
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
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