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inysia [295]
2 years ago
7

Consider the function f(x) = 3x2 + 7x + 2. What is the value of the discriminant? –17253173 How many x-intercepts does this func

tion have? 0123 What are the number of zeros for this function? zero real number solutionsone real number solutiontwo real number solutions
Mathematics
2 answers:
nata0808 [166]2 years ago
8 0

Answer:

two real number solutions

Step-by-step explanation:

Given is a quadratic funciton

f(x) = 3x^2 + 7x + 2\\

To find discriminant

b^2-4ac = 49-4(3)(2) = 25

Tosolve for x intercepts:

PUt y =0

Then since discriminant is a perfect square, we have two rational roots.

They are

\frac{-7+/- 5}{6} =\frac{-1}{3} ,-2

No of zeroes = no of x intercepts =2

Hence answer is

two real number solutions

myrzilka [38]2 years ago
7 0
What is the value of the discriminant?
 For this case, the discriminant will be given by
 b ^ 2 - 4 * a * c
 Where
 b = 7
 a = 3
 c = 2
 substituting
 b ^ 2 - 4 * a * c = (7) ^ 2 - 4 * (3) * (2) = 25
 Therefore the value of the discriminant is 25.
 How many x-intercepts does this function have?
 It has two intercepts with the x axis and can be found by equaling the function to zero. That is to say,
 3x2 + 7x + 2 = 0
 The results will be the interceptions with x.
 What are the number of zeros for this function?
 The number of zeros for this function is
 two real number solutions
 Because it is a quadratic function.
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A million years ago, an alien species built a vertical tower on a horizontal plane. When they returned they discovered that the
velikii [3]

Answer:

\theta=cos^{-1}(\frac{2}{3})

Step-by-step explanation:

We are given that measurements of three points on the ground gave coordinates of (0,0,0),(1,2,0) and (0,2,1)

We have to find the angle by which the tower now deviate from the vertical

We find cross product of <1,2,0> and <0,2,1>

\times =\begin{vmatrix}i&j&k\\1&2&0\\0&2&1\end{vmatrix}

\times =2\hat{i}-\hat{j}+2\hat{k}

Now, we are finding the angle between  \timesand vertical vector <0,0,1>

Angle between two vectors formula

cos\theta=\frac{a.b}{\mid a\mid\cdot\mid b\mid }

Now, using this formula

cos\theta=\frac{2}{1\cdot 3}=\frac{2}{3}

\theta=cos^{-1}(\frac{2}{3})

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6 0
2 years ago
If cos Θ = square root 2 over 2 and 3 pi over 2 &lt; Θ &lt; 2π, what are the values of sin Θ and tan Θ?
KIM [24]

Answer:

The answer is

sin(\theta)=-\frac{\sqrt{2}}{2}

tan(\theta)=-1

Step-by-step explanation:

we know that

tan(\theta)=\frac{sin(\theta)}{cos(\theta)}

sin^{2}(\theta)+cos^{2}(\theta)=1

In this problem we have

cos(\theta)=\frac{\sqrt{2}}{2}

\frac{3\pi}{2}

so

The angle \theta belong to the third or fourth quadrant

The value of sin(\theta) is negative

Step 1

Find the value of  sin(\theta)

Remember

sin^{2}(\theta)+cos^{2}(\theta)=1

we have

cos(\theta)=\frac{\sqrt{2}}{2}

substitute

sin^{2}(\theta)+(\frac{\sqrt{2}}{2})^{2}=1

sin^{2}(\theta)=1-\frac{1}{2}

sin^{2}(\theta)=\frac{1}{2}

sin(\theta)=-\frac{\sqrt{2}}{2} ------> remember that the value is negative

Step 2

Find the value of tan(\theta)

tan(\theta)=\frac{sin(\theta)}{cos(\theta)}

we have

sin(\theta)=-\frac{\sqrt{2}}{2}

cos(\theta)=\frac{\sqrt{2}}{2}

substitute

tan(\theta)=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}

tan(\theta)=-1

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