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Sav [38]
3 years ago
15

Write an equation of the line that passes through points (3, 7) and (5, 11).

Mathematics
1 answer:
dolphi86 [110]3 years ago
3 0

Answer:

B. y – 7 = 2(x – 3)

Step-by-step explanation:

First, find the the <em>rate</em><em> </em><em>of</em><em> </em><em>change</em><em> </em>[<em>slope</em>]:

-y₁ + y₂\-x₁ + x₂ = m

\frac{-7 + 11}{-3 + 5} = \frac{4}{2} = 2

Now, according to the Point-Slope Formula, <em>y - y₁ = m(x - x₁)</em>, all the negative symbols give the OPPOSITE terms of what they really are, so be EXTREMELY CAREFUL inserting the coordinates into the formula with their CORRECT SIGNS.

* Slope = <em>m</em>

I am joyous to assist you anytime.

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Read 2 more answers
Sq is an angle bisector m &lt;qst=2x and m &lt;qsr=3x-10 what is &lt;rst
defon
Answer:
∠RST = 40°

Explanation:
We are given that:
SQ bisects angle RST.
This means that:
∠QST = ∠QSR
2x = 3x - 10
10 = 3x - 2x
x = 10

Therefore:
∠QST = 2x = 2(10) = 20°
∠QSR = 3x - 10 = 3(10) - 10 = 30 - 10 = 20°
Note that both angles are equal.

Now, we can get ∠RST as follows:
∠RST = ∠QST + ∠QSR
∠RST = 20 + 20 = 40°

Hope this helps :)
6 0
3 years ago
Find the interval(s) of upward concavity on this accumulation function.
maxonik [38]

\displaystylef(x)=\int_{0}^{x^2}\sec^2(\sqrt{x})dx \\=\int_{0}^{\sqrt{x^2}}\sec^2(u)\cdot2udu \\=2\int_{0}^{\sqrt{x^2}}\sec^2(u)du \\=2\Big[u\tan(u)-\int\tan(u)du\Big]_{0}^{\sqrt{x^2}} \\=2\Big[u\tan(u)+\ln\Big(\mathrm{abs}(\cos(u))\Big)\Big]_0^x \\=2\Big(x\tan(x)+\dfrac{1}{2}\ln\Big(\cos^2(x)\Big)\Big) \\=2x\tan(x)+\ln(\cos^2(x))

Hope this helps.

6 0
3 years ago
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