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irina [24]
3 years ago
13

How do you solve the following equation: 5-(2x-3)=-8+2x

Mathematics
1 answer:
kirza4 [7]3 years ago
4 0
I believe it would be done like this:
subtract 2x-8 from both sides. then use cubric formula.
answer should be  x= 1.688242

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Plzz help with this one problem ASAP
shtirl [24]

Answer:

i dont know what the problem is

Step-by-step explanation:

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3 years ago
One method we can use to solve some polynomial equations
kotykmax [81]

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7 0
3 years ago
If f(x)=5x-25 and g(x) = 1/5x+5 which expression could be used to verify g(×) is the inverse of f(×)
Gelneren [198K]

Answer:

The  expression could be used to verify g(x) is the inverse of f(x) is letter b which is 1/5(5x-25)+5

Step-by-step explanation:


4 0
3 years ago
Read 2 more answers
The point (1/3,1/4) lies on the terminal said of an angle. Find the exact value of the six trig functions and explain which func
katrin2010 [14]

Answer:

sine and cosec are inverse of each other.

cosine and sec are inverse of each other.

tan and cot are inverse of each other.

Step-by-step explanation:

Given point on terminal side of an angle (\frac{1}{3},\frac{1}4).

Kindly refer to the attached image for the diagram of the given point.

Let it be point A(\frac{1}{3},\frac{1}4)

Let O be the origin i.e. (0,0)

Point B will be (\frac{1}{3},0)

Now, let us consider the right angled triangle \triangle OBA:

Sides:

Base, OB = \frac{1}{3}\\Perpendicular, AB = \frac{1}{4}

Using Pythagorean theorem:

\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow OA^{2} = OB^{2} + AB^{2}\\\Rightarrow OA^{2} = \frac{1}{3}^{2} + \frac{1}{4}^{2}\\\Rightarrow OA = \sqrt{\frac{1}{3}^{2} + \frac{1}{4}^{2}}\\\Rightarrow OA = \sqrt{\frac{4^2+3^2}{3^{2}.4^2 }}\\\Rightarrow OA = \frac{5}{12}

sin \angle AOB = \dfrac{Perpendicular}{Hypotenuse}

\Rightarrow sin \angle AOB = \dfrac{\frac{1}{4}}{\frac{5}{12}}\\\Rightarrow sin \angle AOB = \dfrac{3}{5}

cos\angle AOB = \dfrac{Base}{Hypotenuse}

\Rightarrow cos \angle AOB = \dfrac{\frac{1}{3}}{\frac{5}{12}}\\\Rightarrow cos\angle AOB = \dfrac{4}{5}

tan\angle AOB = \dfrac{Perpendicular}{Base}

\Rightarrow tan\angle AOB = \dfrac{3}{4}

cosec \angle AOB = \dfrac{Hypotenuse}{Perpendicular}

\Rightarrow cosec\angle AOB = \dfrac{5}{3}

sec\angle AOB = \dfrac{Hypotenuse}{Base}

\Rightarrow sec\angle AOB = \dfrac{5}{4}

cot\angle AOB = \dfrac{Base}{Perpendicular}

\Rightarrow cot\angle AOB = \dfrac{4}{3}

3 0
3 years ago
Is the answer to this question correct?
Komok [63]
It looks about right to me.
5 0
3 years ago
Read 2 more answers
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