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2 years ago

Which of the lines shown above has a negative slope? (1)A (2)B (3)C (4)D (5)E

Mathematics

2 answers:

LuckyWell [14K]2 years ago

7 0

Answer:

Step-by-step explanation:

B is the only one going own from left to right 2 of them are neutral and 2 are positive

horrorfan [7]2 years ago

4 0

Answer:

the correct answer is B

Step-by-step explanation:

hope this helps

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HELP WITH THESE TWO MATH QUESTIONS PLEASE!!!<br> ASAP!

Brilliant_brown [7]

The first picture would be choice answer c because the number of x's represent how many birds wings are this size. The number on the tic mark represents the size of the wing span. So you would take the number of x's times the tic mark represented under it. There are 3 birds with 3 1/8 wingspan so it wold be 3 1/8 *3. Hoped this helps for the first one!

8 0

2 years ago

Does this graph represent a function? Why or why not?

In-s [12.5K]

B. No, because it fails the vertical line test.

8 0

3 years ago

If the mean of the data below is 3.8, find m<br>x 1 2 3 4 5 6<br>f 1 4 7 m 6 3

zloy xaker [14]

Answer:

m = 1.8

Step-by-step explanation:

Given data is as follows :

x 1 2 3 4 5 6

f 1 4 7 m 6 3

We need to find the value of m if the mean of the data is 3.8.

Mean = sum of observations/no. of observations

So,

$3.8=\dfrac{1+4+7+m+6+3}{6}\\\\22.8=21+m\\\\22.8-21=m\\\\m=1.8$

So, the value of m is equal to 1.8.

8 0

3 years ago

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

3 years ago

Create a unique parabola in the pattern f(x) = (x − a)(x − b)

MissTica

Answer:

f(x)=(x-6)(x-(-7))

Step-by-step explanation:

Foiled:

0 = x + x - 42

5 0

2 years ago