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r-ruslan [8.4K]
3 years ago
12

How many real solutions does a quadratic equation have if its discriminant is zero?

Mathematics
1 answer:
Flura [38]3 years ago
8 0
It is one answer with multiplicity 2
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A cube with an edge length s has a volume of 8 cubic units. What is the length of s?
GrogVix [38]

Answer:

Perimeter on one face =4× side =8

Hence the side of the cube is 2 units.

So the volume of the cube is (side)^3=(2)^3=8 cubic units

Step-by-step explanation:

3 0
3 years ago
A researcher wished to compare the effect of two stepping heights (low and high) on heart rate in a step-aerobics workout. A col
Alinara [238K]

Answer:

The answer is "between 0.10 and 0.05".

Step-by-step explanation:

For sample 1:

n_1 = 25\\\\\bar{x_1} = 90.00\\\\ s_1 = 9\\\\

For sample 2:

n_2 = 25\\\\ \bar{x_2} = 95.08\\\\ S_2 = 12\\\\

Calculating the null and alternative hypotheses:

H_O : \mu_1 = \mu_2 \\\\ H_Q : \mu_1  

Calculating the test statistic:

Z= \frac{90-95.08}{ \sqrt{\frac{9^2}{25}+\frac{12^2}{25}}} \\\\ =\frac{-5.08}{ \sqrt{\frac{81}{25}+\frac{144}{25}}}\\\\=\frac{-5.08}{ \sqrt{\frac{81+144}{25}}}\\\\=\frac{-5.08}{ \sqrt{\frac{225}{25}}}\\\\=\frac{-5.08}{ \frac{15}{5}}\\\\=\frac{-5.08}{3}\\\\= -1.693

Calculating the conservative degrees of freedom:

DF = min (n_1 -1, n_2 - 1) = min(25-1, 25-1) = 24

by using Excel the p-value for left tailed test and for test statistic will be -1.693 with DF = 24.

\to p-value = TDIST(1.693, 24, 1) = 0.0517\ \ \ \  i.e, \bold{0.05 < p-value < 0.10}

5 0
3 years ago
Find the volume of this figure. Round your answer to the nearest hundredth, if necessary.
solmaris [256]

Answer:

V = 3617.28 m^3

Step-by-step explanation:

Volume of  a cone

V = 1/3 πr^2h

Where r = 24/2 = 12 m

h = 24 m

Plug in

V = 1/3 (3.14) (12^2)(24)

V = 3617.28 m^3

5 0
3 years ago
What is the surface area of this cylinder?
Cloud [144]

Answer:

Lateral surface area: L = 50.265482457437 yd^2

Top surface area T = 12.566370614359 yd^2

Base surface area B = 12.566370614359 yd^2

Total surface area A = 75.398223686155 yd^2

Step-by-step explanation:

5 0
4 years ago
Determine whether the set of vectors <img src="https://tex.z-dn.net/?f=%20v_%7B1%3D%283%2C2%2C1%29%2C%20v_%7B2%7D%20%3D%28-1%2C-
Korolek [52]
Since each vector is a member of \mathbb R^3, the vectors will span \mathbb R^3 if they form a basis for \mathbb R^3, which requires that they be linearly independent of one another.

To show this, you have to establish that the only linear combination of the three vectors c_1\mathbf v_1+c_2\mathbf v_2+c_3\mathbf v_3 that gives the zero vector \mathbf0 occurs for scalars c_1=c_2=c_3=0.

c_1\begin{bmatrix}3\\2\\1\end{bmatrix}+c_2\begin{bmatrix}-1\\-2\\-4\end{bmatrix}+c_3\begin{bmatrix}1\\1\\-1\end{bmatrix}=(0,0,0)\iff\begin{bmatrix}3&-1&1\\2&-2&1\\1&-4&-1\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}

Solving this, you'll find that c_1=c_2=c_3=0, so the vectors are indeed linearly independent, thus forming a basis for \mathbb R^3 and therefore they must span \mathbb R^3.
4 0
3 years ago
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