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3 years ago

Im need help on this question

Mathematics

1 answer:

dimulka [17.4K]3 years ago

4 0

You were right, its the second choice CA=CB+BA

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Given g(x) = ln x, is it true that if u>0 and v>0 g(u) = 2g(v), then u=v?

Leokris [45]

Answer:

False

Step-by-step explanation:

<h3>Given</h3>

g(x) = ln x

g(u) = 2g(v)

<h3>Solution</h3>

g(u) = ln u
2g(v) = 2ln v

<u>Comparing</u>

g(u) = 2 g(v)
ln u = 2ln v
ln u = ln v²
u = v²

This is not same as u = v, so the answer is false

3 0

3 years ago

Read 2 more answers

Consider the following pair of equations:

frozen [14]

X+4=y=-2x-2
x+4=-2x-2
add 2x to both side
3x+4=-2
minus 4 both sides
3x=-6
divide 3
x=-2

sub back
y=x+4
y=-2+4
y=2

(x,y)
(-2,2)

3 0

3 years ago

The formula for the area of a regular polygon is A=1/2ap. What is the equation solved for a?

Galina-37 [17]

$A= \frac{1}{2}ap \ \to \ 2A =ap \ \to \ \boxed{a= \frac{2A}{p}}$

5 0

4 years ago

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A class has 21 girls and 12 boys. what is the probability that a boy's name is drawn at random

belka [17]

There are a total of 33 students in the class (21+12=33). There are 12 boys in the class so the chance of drawing a boys name would be 12/33 which reduces to 4/11. This means you will draw a boys name 4 times for every 11 draws. To find the percent probability, just divide...

4/11 =.3636

multiply by 100 to convert to a percentage ( move decimal to the right 2 places).

.3636 = 36.36%

There is a 36.36% chance that a boys name is drawn.

3 0

3 years ago

A piece of wire 19 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

mr Goodwill [35]

Answer: 8.26 m

Step-by-step explanation:

$Let s be the length of the wire used for the square. \\Let $t$ be the length of the wire used for the triangle. \\Let $A_{S}$ be the area of the square. \\Let ${A}_{T}}$ be the area of the triangle. \\One side of the square is $\frac{s}{4}$ \\Therefore,we know that,$$A_{S}=\left(\frac{s}{4}\right)^{2}=\frac{s^{2}}{16}$

$The formula for the area of an equilateral triangle is, $A=\frac{\sqrt{3}}{4} a^{2}$ where $a$ is the length of one side,And one side of our triangle is $\frac{t}{3}$So,We know that,$$A_{T}=\frac{\sqrt{3}}{4}\left(\frac{t}{3}\right)^{2}$$We have to find the value of "s" such that,$\mathrm{s}+\mathrm{t}=19$ hence, $\mathrm{t}=19-\mathrm{s}$And$$A_{S}+A_{T}=A_{S+T}$

$$$Therefore,$$\begin{aligned}&A_{T}=\frac{\sqrt{3}}{4}\left(\frac{(19-s)}{3}\right)^{2}=\frac{\sqrt{3}(19-s)^{2}}{36} \\&A_{T+S}=\frac{s^{2}}{16}+\frac{\sqrt{3}(19-s)^{2}}{36}\end{aligned}$

$Differentiating the above equation with respect to s we get,$$A^{\prime}{ }_{T+S}=\frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}$$Now we solve $A_{S+T}^{\prime}=0$$$\begin{aligned}&\Rightarrow \frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}=0 \\&\Rightarrow \frac{s}{8}=\frac{\sqrt{3}(19-s)}{18}\end{aligned}$$Cross multiply,$$\begin{aligned}&18 s=8 \sqrt{3}(19-s) \\&18 s=152 \sqrt{3}-8 \sqrt{3} s \\&(18+8 \sqrt{3}) s=152 \sqrt{3} \\&s=\frac{152 \sqrt{3}}{(18+8 \sqrt{3})} \approx 8.26\end{aligned}$

$The domain of $s$ is $[0,19]$.So the endpoints are 0 and 19$$\begin{aligned}&A_{T+S}(0)=\frac{0^{2}}{16}+\frac{\sqrt{3}(19-0)^{2}}{36} \approx 17.36 \\&A_{T+S}(8.26)=\frac{8.26^{2}}{16}+\frac{\sqrt{3}(19-8.26)^{2}}{36} \approx 9.81 \\&A_{T+S}(19)=\frac{19^{2}}{16}+\frac{\sqrt{3}(19-19)^{2}}{36}=22.56\end{aligned}$

$$$Therefore, for the minimum area, $8.26 \mathrm{~m}$ should be used for the square$

8 0

2 years ago