Can someone plz help would be nice

Mathematics

1 answer:

Tems11 [23]2 years ago

6 0

DONT GO TO THAT LINK ITS A VIRUS AND ITS A BOT

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What are the missing coordinates? My teacher said it was (h-m,n), but I’m saying it is (h+m,n). Could anyone help?

antiseptic1488 [7]

Answer:

You are right, the coordinates are:

$(h+m,n)$

Explanation:

Here is how you can proove that you are right.

Note that m is the x-coordinate of the point $(m,n)$ . Since this point is to the left of the y-axis, it is a negative number, but the negative sign is included in $m$ .

The distance from $(m,n)$ to the y-axis is not $m$ but $-m$ .

That is the same distance from the point with the missing coordinates to the point (h,0).

Then, the missing x-coordinate is $h-(-m)=h+m.$

Of course, there is not doubt about the y-coordinate: it is n, because the point with the missing coordinates is at the same height as (m,n).

Hence, the missing coordinates are x = h+m, and y = n: (h + m, n).

3 0

2 years ago

the inequality |t-57|<7 describes the range of monthly average temperature, T, in Degrees Fahrenheit, for san Francisco, Cali

Y_Kistochka [10]

Solution is: $50\leq T\leq 64$ and it means, the minimum temperature is 50° Fahrenheit and the maximum temperature is 64° Fahrenheit.

Explanation

Given inequality is: $|T-57|\leq 7$

As this is absolute inequality, so we will get two different inequalities-- one for positive and another for negative. So.....

$T-57\leq 7\\ \\ T\leq 7+57\\ \\ T\leq 64$

and

$-(T-57)\leq 7\\ \\ -T+57\leq 7\\ \\ -T\leq 7-57\\ \\ -T\leq -50\\ \\ T\geq 50$

So, the final combined solution will be: $50\leq T\leq 64$

It means that the minimum temperature for San Francisco, California is 50° Fahrenheit and the maximum temperature is 64° Fahrenheit.

7 0

2 years ago

The 3rd and 6th term of a geometric progression are 9/2 and 243/16 respectively find the first term, common ratio, seventh term

sashaice [31]

Answer:

Hello,

Step-by-step explanation:

$Let\ (u_n)\ the\ geometric\ progression.\\\\r\ is\ the\ common\ ratio.\\\\u_3=u_0*r^3\\u_6=u_0*r^6\\\\\dfrac{u_6}{u_3} =r^3=\dfrac{\frac{243}{16} }{\frac{9}{2} } =\dfrac{27}{8} =(\frac{3}{2} )^3\\\\\boxed{r=\dfrac{3}{2} }\\\\\\u_3=u_1*r^2 \Longrightarrow\ u_1=\dfrac{u_3}{r^2} =\dfrac{\frac{9}{2} }{(\frac{3}{2^2}) } =2\\\\\\u_7=u_6*\dfrac{3}{2} =\dfrac{729}{32}$