Evaluate 1+3+5+...+25

Find the length of UW(with a line over it) if W is between U and V, UV = 16.8 centimeters, and VW = 7.9 centimeters.

stealth61 [152]

Answer:

8.9 cm

Step-by-step explanation:

We have
$\overline {UV} = 16.8 cm\\\\\overline {WV} = 7.9 cm\\\\ Since all lines are scalars, \overline{WV} = \overline{VW}\\\textrm{We see that \overline{UW} + \overline{WV} = \overline{UV}\\\\\overline {UW} \\ = \overline {UV} - \overline {WV} = 16.9 - 7.9 = 8.9 cm}$

7 0

1 year ago

Please help im confused on the last question on my warm up!

geniusboy [140]

Answer:

Choice B: Only (-1,-5)

Step-by-step explanation:

With this one, first the equation needs to be reduced into point-slope format.

y+5=2(x+1)

Distributive Prop.

y+5=2x+2

Carry over the 5 to the opposite side.

y=2x-3

Now, with the point slope form, it makes it easier to plug in the question choices to find which if any are correct.

Choice A: X is 5; Y is 10

y=2x-3

y=2(5)-3

y=10-3

y=7

This does not give a y value of 10, so it is false.

Choice B: X is -1; Y is -5

y=2(x)-3

y=2(-1)-3

y= -2-3

y= -5

This does give a y value of -5, so it is true. Choices C and D are therefore false as well, since only Choice B works.

3 0

3 years ago

Law of Syllogism.

stiks02 [169]

Answer:

true

Step-by-step explanation:

if it is 10,20,30,40etc...

10÷5=2

20÷5=10

30÷5=15

40÷5=20

3 0

3 years ago

Write and solve a system of equations for the situation described below. Define your variables and write your solution as a sent

Delicious77 [7]

N = nickels and q = quarters

0.05n + 0.25q = 1.90
n = 2q + 3

0.05(2q + 3) + 0.25q = 1.90
0.10q + 0.15 + 0.25q = 1.90
0.10q + 0.25q = 1.90 - 0.15
0.35q = 1.75
q = 1.75/0.35
q = 5 ......there are 5 quarters

n = 2q + 3
n = 2(5) + 3
n = 10 + 3
n = 13 <=== there are 13 nickels

6 0

3 years ago

Find the domain of the function y = 3 tan(23x)

solmaris [256]

Answer:

$\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

In other words, the $x$ in $f(x) = 3\, \tan(23\, x)$ could be any real number as long as $x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)$ for all integer $k$ (including negative integers.)

Step-by-step explanation:

The tangent function $y = \tan(x)$ has a real value for real inputs $x$ as long as the input $x \ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

Hence, the domain of the original tangent function is $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

On the other hand, in the function $f(x) = 3\, \tan(23\, x)$ , the input to the tangent function is replaced with $(23\, x)$ .

The transformed tangent function $\tan(23\, x)$ would have a real value as long as its input $(23\, x)$ ensures that $23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

In other words, $\tan(23\, x)$ would have a real value as long as $x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right)$ .

Accordingly, the domain of $f(x) = 3\, \tan(23\, x)$ would be $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

4 0

3 years ago