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docker41 [41]
3 years ago
6

V² + 2v-8=0 this question accords using the quadratic formula as equation​

Mathematics
2 answers:
yaroslaw [1]3 years ago
8 0
Hope this helps with your answer.

kirill [66]3 years ago
6 0
That’s how to solve it
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432,2=26<br>275,3=66<br>342,5=41<br>123,4=20<br>then <br>543,2=?​
Readme [11.4K]

Answer:

I think 543,2=24.

hope it helps

5 0
3 years ago
The weight of oranges growing in an orchard is normally distributed with a mean weight of 6.5 oz. and a standard deviation of 1.
sergey [27]
Add 3 standard deviations above and below the mean to get the range in which 99.7% of the data in a normal distribution will fall
6.5 + 4.5 = 11
6.5 - 4.5 = 2
So 2 to 11 ounces would be the interval
6 0
3 years ago
Please answer ASAP!! Thanks
charle [14.2K]
-3 carpark (c)
When you go up 3 levels, you are at level 3 and then you go down 6 which is -3
8 0
3 years ago
Pleas help me find the composition of goh and it’s domain
Ksivusya [100]

Solution:

The function is given below as

\begin{gathered} g(x)=\frac{x+4}{x-1} \\ h(x)=2x-1 \end{gathered}

To figure out

(goh)(x)

To do this , we will substitute x= 2x-1 in g(x)

\begin{gathered} g(x)=\frac{x+4}{x-1} \\ g(h)(x)=\frac{2x-1+4}{2x-1-1} \\ g(h)(x)=\frac{2x+3}{2x-2} \end{gathered}

Hence,

The composte function will be

(goh)(x)=\frac{2x+3}{2x-2}

Step 2:

To figure out the domain,

In mathematics, the domain of a function is the set of inputs accepted by the function.

Hence,

The domain of the function is

\begin{bmatrix}\mathrm{Solution:}\:&\:x1\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:1\right)\cup \left(1,\:\infty \:\right)\end{bmatrix}

5 0
1 year ago
Cot? 2x + cos2x + sin’ 2x = csc? 2x<br> Need help with 40
kkurt [141]

Answer:

\rule{100mm}{0.1mm}

    \cot^2(2x)+\cos^2(2x)+\sin^2(2x)=\csc^2(2x)

\dfrac{\cos^2(2x)}{\sin^2(2x)} +[(\cos^2(2x)+\sin^2(2x)]=\csc^2(2x)

                               \dfrac{\cos^2(2x)}{\sin^2(2x)} +1=\csc^2(2x)

                    \dfrac{\cos^2(2x)+\sin^2(2x)}{\sin^2(2x)}=\csc^2(2x)

                                    \dfrac{1}{sin^2(2x)} =\csc^2(2x)

                                     \boxed{\csc^2(2x)=\csc^2(2x)} ✔

\rule{100mm}{0.1mm}

4 0
3 years ago
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