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Likurg_2 [28]
3 years ago
11

Who uses iPhone x silver for editing on vs or am I the only one._.

Mathematics
2 answers:
sineoko [7]3 years ago
8 0

Answer:

I use it too! :P

Sunny_sXe [5.5K]3 years ago
4 0

Step-by-step explanation:

I don't use it though

maybe someday

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Simplify (3x+2)(x-7)
vovangra [49]
The answer is
3 {x}^{2}  - 19x - 14
first you must multiply or use the FOIL method (First, Outer, Inner, Last)
First you multiply the first terms
(3x)(x)
this gives you 3x^2, then you multiple the outer terms
( - 7)(3x)
this gives you -21x, then you multiply the inner terms
(2)(x)
that gives you 2x, finally you multiply the last terms
(2)( - 7)
that gives you 14, now you combine like terms
- 21x + 2x =  - 19x
now you put all the terms together and end up with your solution
3 {x}^{2}  - 19x - 14

8 0
3 years ago
Read 2 more answers
Plot the x- and y-intercepts to graph the equation. y=−12x−3 could you guys plot it on a graph? if so thank you whichever one is
tamaranim1 [39]

Answer:

(-4, 0)  (8, 0)

Step-by-step explanation:

I did the K12 test.

3 0
2 years ago
If f(x)=4/x+2 and g is the inverse of f, then g'(10)=​
evablogger [386]

Answer:

The value of g'(10)=\frac{(-1)}{16}

Step-by-step explanation:

Given function is f(x)=\frac{4}{x} + 2

Take f(x)=y

y=\frac{4}{x} + 2

Subtract 2 from both side.

y-2=\frac{4}{x}

x=\frac{4}{y-2}

The inverse of f(x) is written as \frac{4}{y-2}

It is said as g is inverse of f

g(y)=\frac{4}{y-2}[/tex]

g(y)=\frac{4}{y-2}

g(10)=\frac{4}{10-2}

g(10)=\frac{1}{2}

Differentiating both sides we get,

g'(y)=\frac{4(-1)}{(y-2)^{2}}+0

g'(y)=\frac{(-4)}{(y-2)^{2}}

To find g'(10)=\frac{(-4)}{(10-2)^{2}}

g'(10)=\frac{(-1)}{16}

5 0
3 years ago
In ABC, a = 13,b=21, and c = 27. Find mA.
coldgirl [10]

Answer:

Step-by-step explanation:

a = 13  ; b = 21  ; c = 27

S = (a + b + c)/2

   = \dfrac{13+21+27}{2}\\\\=\dfrac{61}{2}\\\\= 30.5

s- a = 30.5 - 13 = 17.5

s - b = 30.5 - 21 = 9.5

s  -c = 30.5 - 27 = 3.5

Area = \sqrt{s(s-a)(s-b)(s-c)}

        = \sqrt{30.5*17.5*9.5*3.5}\\\\=\sqrt{17747.1875} \\\\=133.21\\\\=133

\dfrac{cbSin \ A}{2}=133\\\\\dfrac{27*21*Sin A}{2}=133\\\\Sin \ A = \dfrac{133*2}{27*21}\\\\Sin \ A = 0.4691\\\\A = Sin^{-1} \ 0.4691\\\\\\

A = 27°

4 0
3 years ago
Prove: csc theta/sin theta - cot theta/tan theta<br><br>Answer: 1
Marysya12 [62]
\bf \cfrac{cos(\theta )}{sin(\theta )}-\cfrac{cot(\theta )}{tan(\theta )}=1\implies \cfrac{cos(\theta )}{sin(\theta )}-\cfrac{\frac{cos(\theta )}{sin(\theta )}}{\frac{sin(\theta )}{cos(\theta )}}=1&#10;\\\\\\&#10;\cfrac{cos(\theta )}{sin(\theta )}-\cfrac{cos(\theta )}{sin(\theta )}\cdot \cfrac{cos(\theta )}{sin(\theta )}=1&#10;\implies &#10;\cfrac{cos(\theta )}{sin(\theta )}-\cfrac{cos^2(\theta )}{sin^2(\theta )}=1

\bf \cfrac{cos(\theta )sin(\theta )~~-~~cos^2(\theta )}{sin^2(\theta )}=1\implies cos(\theta )sin(\theta )-cos^2(\theta )\ne sin^2(\theta )


now, if we move the second fraction over, we'd get the equation of,

\bf \cfrac{cos(\theta )}{sin(\theta )}-\cfrac{cot(\theta )}{tan(\theta )}=1\implies \cfrac{cos(\theta )}{sin(\theta )}=1+\cfrac{cot(\theta )}{tan(\theta )}


now, check the picture below, they are definitely not equal to one another.

5 0
2 years ago
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