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2 years ago

Which figure has reflection symmetry?

Mathematics

2 answers:

lara31 [8.8K]2 years ago

8 0

Answer:

the first option

Step-by-step explanation:

if you were to fold it in half it would be the same shape without anything poking out of the sides. one side reflects the other side

seraphim [82]2 years ago

5 0

The bottom left figure

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What is the probability that a two-digit number selected at random has a tens digit less than its units digit

Reika [66]

The probability that a two-digit number selected at random has a tens digit less than its units digit is 0.2667 (4/15).

$A=24B=90\\\\P=A/B\\\\P=24/90\\p=4/15$

There are 90 two-digit numbers (99-9). Of these, six numbers are divisible by 15 (15, 30, 45, 60, 75, 90). This is also divisible by 5. Therefore, the preferred case is 30-6 = 24. Therefore, the required probability is 24/90 = 4/15.

The probability of an event can be calculated by simply dividing the number of favorable results by the total number of possible results using a probabilistic expression. Whenever you are uncertain about the outcome of an event, you can talk about the probability of a particular outcome, that is, its potential.

Learn more about probability here: brainly.com/question/24756209

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2 years ago

For 32 = 5X + 2, what is the first step in solving for X?

nikitadnepr [17]

Answer:

-2 from both sides

Step-by-step explanation:

32=5x+2

-2 -2

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30=5x

__=__

5 5

6=x

4 0

3 years ago

What is the answer? Plz help

zloy xaker [14]

Answer:

5:15; 5 cups of juice and 15 cups of lemonade

Step-by-step explanation:

when simplified, the ratio is 1:3; so the first number multiplied by 3 equals the 2nd number

other answers: 6:18, 7:21, 8:24

8 0

3 years ago

Read 2 more answers

The endpoints of CD are C(–8, 4) and D(6, –6). What are the coordinates of point P on CD such that P is the length of the line s

GalinKa [24]

If you graph the end points C and D then graph the 4 points at the end it is difficult to tell which points are on CD without a line.
Using the endpoints find the slope (change in y/ change in x) then substitute a point in to find the intercept.
Slope = (-6-4)/(6- -8) = -5/7
Intercept equation (-6) = -5/7 (6) + b
b = -1.71428571429
Graphing the line shows only 2 points on the line (–2.75, 0.25) and <span>(0.75, –2.25)
I am confused by the part, "</span><span>P is the length of the line segment from D". Were you given a length P to help you determine which point. Using the distance formula to find the length from each point to D doesn't help determine which one is best with the information you have given. The image shows the distances I calculated and the graphed points.
I hope this helps!</span>

5 0

3 years ago

Read 2 more answers

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