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3 years ago

Groks mother weighed

Mathematics

2 answers:

Kruka [31]3 years ago

6 0

The answer is 66 your welcome

shusha [124]3 years ago

4 0

Answer:

62 Kg

Step-by-step explanation:

Let g = weight of Grok last year in kg

m = weight of Grok's mother last year in kg

m = 3g

m-4 = 2(g+9)

since m = 3g

(3g)-4 = 2(g+9)

3g -4 = 2g + 18

3g-2g = 18+4

g = 22

m = 3*22

= 66

Note that:

g+9 = weight of Grok now

m-4 = weight of Grok's mother now

Grok weighs 31 Kg now

Grok's mother weighs 62 Kg now

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How

andrey2020 [161]

Answer: 7800

Step-by-step explanation:

As English alphabet has 26 letters.

So , Choices for First character in the code = 26

The choices for digits = 10 (From 0 to 9)

So , choices for second character = 10 (Middle two characters must be numbers)

choices for third character = 10 (Middle two characters must be numbers)

Choices for last character = 3 (the last character must be a 1, 3, or 7)

By Fundamental principle of counting , the total number of four character codes can be generated =

(Choices for First character)x (choices for digits) x(choices for digits) x(Choices for last character)

= (26) (10)(10)(3) =7800

Hence, the total number of character codes can be generated= 7800

6 0

3 years ago

Que is on pic.i can't able to type in text.

ad-work [718]

It's not difficult to compute the values of $A$ and $B$ directly:

$A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}$
$A=\tan^{-1}(\sin\theta)-\dfrac\pi4$

$B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt$
$B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}$
$B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2$

Let's assume $0$ , so that $|\csc\theta|=\csc\theta$ .

Now,

$\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}$
$\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}$
$\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)$
$\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)$

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in $A$ and $B$ .

3 0

2 years ago

I need help asap! 25 points

alukav5142 [94]

Answer:bc thats not the number included in the problem

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

An interval has the notation (-6,-2). find the midpoint

notka56 [123]

<span>(6+8)/2 = 14/2 = 7 hope this helps

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