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4 years ago

PLEASE HELP!! Figure ABC is reflected about the x-axis to obtain figure A’B’C’:

Mathematics

2 answers:

Feliz [49]4 years ago

7 0

Answer:

The measure of angle D is equal to the measure of angle A′.

Step-by-step explanation:

Radda [10]4 years ago

5 0

The answer I believe is D.

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Unit rates can be helpful with creating a ratio table and plotting the equivalent ratios on a coordinate plane. Explain why. PLE

uysha [10]

Answer:

Unit rate is often a useful means for comparing ratios and their associated rates when measured in different units. The unit rate allows us to compare varying sizes of quantities by examining the number of units of one quantity per one unit of the second quantity. This value of the ratio is the unit rate.

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B6%7D%7B25%7D%20%3D%20%20%5Cfrac%7Bd%7D%7B30%7D%20" id="TexFormula1" title=" \frac

GarryVolchara [31]

Answer:

For $\frac{6}{25} = \frac{d}{30}$ , d = 7.2

Step-by-step explanation:

Here, the given expression is $\frac{6}{25} = \frac{d}{30}$

To find the value of the variable d :

$\frac{6}{25} = \frac{d}{30} \implies d = \frac{6}{25} \times 30$

or, $d = \frac{180}{25} = \frac{36}{5 } = 7.2$

Hence, for $\frac{6}{25} = \frac{d}{30}$ , d = 7.2

5 0

3 years ago

Events AA and BB are independent. P(A\ \mathrm{and}\ B)=0.25P(A and B)=0.25 Enter possible probabilities for events AA and BB.

xenn [34]

Answer:0.5, 0.5

Step-by-step explanation:

Given

$P(A\cap B)=0.25$

Also, A and B are independent events. So, we can write that

$P(A\cap B)=P(A)\times P(B)$

$0.25=P(A)\times P(B)$

So, the two most possible value of $P(A)\ \text{and}\ P(B)$ are $0.5\ \text{and}\ 0.5$

8 0

3 years ago

PLEASE HELP ME!!!! THANK YOU SO SO SO MUCH.

polet [3.4K]

It would be A because everything is doubled

4 0

4 years ago

Read 2 more answers

Find f(g(x)). State the domain of the composite function.<br> f(x) = 3x-2/x+1<br> g(x) = x+5/2x-3

AnnyKZ [126]

Answer:

ℝ - {(-2/3),(3/2)}

Step-by-step explanation:

We want the domain of f(g(x)). So, firstly, we have to find the domain for g(x) and, then, for f(g(x)).

- Domain of g(x): Since the expression is a fracion, we must exclude the values of x that make null the denominator. Hence,

$g(x)=\dfrac{x+5}{2x-3}\Longrightarrow 2x-3\neq 0\iff \boxed{x\neq\dfrac{3}{2}}$

- Domain of f(g(x)): We'll find its expression:

$f(x) = \dfrac{3x-2}{x+1}\\\\f(g(x)) = \dfrac{3g(x)-2}{g(x)+1}\\\\f(g(x)) = \dfrac{3\cdot\dfrac{x+5}{2x-3}-2}{\dfrac{x+5}{2x-3}+1}=\dfrac{~~~\dfrac{3(x+5)-2(2x-3)}{2x-3}~~~}{\dfrac{(x+5)+(2x-3)}{2x-3}}\\\\f(g(x)) =\dfrac{3(x+5)-2(2x-3)}{(x+5)+(2x-3)}=\dfrac{3x+15-4x+6}{x+5+2x-3}\\\\\boxed{f(g(x)) =\dfrac{21-x}{3x+2}}$

Now, once again, we have to exclude the values of x that make the denominator equals to zero. Thus,

$f(g(x)) =\dfrac{21-x}{3x+2}\Longrightarrow 3x+2\neq0\iff \boxed{x\neq-\dfrac{2}{3}}$

Lastly, we may write the domanin of f(g(x)):

$D(f(g(x)) = \left]-\infty,-\dfrac{2}{3}\right[\cup\left]-\dfrac{2}{3},\dfrac{3}{2}\right[\cup\left]\dfrac{3}{2},\infty\right[$

or, just writing in a shorter way:

$\boxed{D(f(g(x)) = \mathbb{R}-\left\{-\dfrac{2}{3},\dfrac{3}{2}\right\}}$

7 0

3 years ago