(3x^2-15d) - (-10x^2-4d)

Mathematics

2 answers:

ElenaW [278]3 years ago

6 0

<span>(3x^2-15d) - (-10x^2-4d) = 13x^2-11d

Hope this helps!!
</span>

kirill [66]3 years ago

4 0

Yehhhh man whattt hee said im just trynnna get more points

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The midpoint of RS¯¯¯¯¯¯¯ is M. Use the given endpoint R(23, 14) and midpoint M(13, 8) to find the coordinates of the other endp

Oksanka [162]

The coordinates of other endpoint S is (3, 2)

<h3><u>Solution:</u></h3>

Given that midpoint of RS is M

Given endpoint R(23, 14) and midpoint M(13, 8)

To find: coordinates of the other endpoint S

<em><u>The formula for midpoint is given as:</u></em>

For a line containing containing two points $(x_1, y_1)$ and $(x_2, y_2)$ midpoint is given as:

$m(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Here in this problem,

m(x, y) = (13, 8)

$(x_1, y_1) = (23, 14)\\\\(x_2, y_2) = ?$

Substituting the given values in above formula, we get

$(13,8)=\left(\frac{23+x_{2}}{2}, \frac{14+y_{2}}{2}\right)$

Comparing both the sides we get,

$\begin{aligned}&13=\frac{23+x_{2}}{2} \text { and } 8=\frac{14+y_{2}}{2}\\\\&26=23+x_{2} \text { and } 16=14+y_{2}\\\\&x_{2}=3 \text { and } y_{2}=2\end{aligned}$

Thus the coordinates of other endpoint S is (3, 2)

8 0

3 years ago

URGENT HELP ME PLEASE

Trava [24]

Answer:

(a) $\log_3(\dfrac{81}{3})=3$

(b) $\log_5(\dfrac{625}{25})=2$

(d) $\log_4(\dfrac{64}{16})=1$

(e) $\log_6(36^4)=8$

(f) $\log(100^3)=6$

Step-by-step explanation:

Let as consider the given equations are $\log_3(\dfrac{81}{3})=?,\log_5(\dfrac{625}{25})=?,\log_2(\dfrac{64}{8})=?,\log_4(\dfrac{64}{16})=?,\log_6(36^4)=?,\log(100^3)=?$ .