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tresset_1 [31]
3 years ago
12

What is 5 X 1 over 3/5

Mathematics
2 answers:
Fynjy0 [20]3 years ago
5 0
I got 8 1/3 (8.3 is what it would round to)
dybincka [34]3 years ago
3 0

Answer:

0.3

Step-by-step explanation:

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What pattern would to you expect when two variables are related by a linear model with a positive slope
olganol [36]
There isnt one its has to be consistent but honestly idk
4 0
2 years ago
The vertices of a triangle ABC are A(7, 5), B(4, 2), and C(9, 2). What is measure of angle ABC? 30° 45° 56.31° 78.69°
GenaCL600 [577]

The measure of angle ABC is 45°

<em><u>Explanation</u></em>

Vertices of the triangle are:   A(7, 5), B(4, 2), and C(9, 2)

According to the diagram below....

Length of the side BC (a) =\sqrt{(4-9)^2+(2-2)^2}= \sqrt{25}= 5

Length of the side AC (b) = \sqrt{(7-9)^2 +(5-2)^2}= \sqrt{4+9}=\sqrt{13}

Length of the side AB (c) = \sqrt{(7-4)^2 +(5-2)^2} =\sqrt{9+9}=\sqrt{18}

We need to find ∠ABC or ∠B . So using <u>Cosine rule</u>, we will get...

cosB= \frac{a^2+c^2-b^2}{2ac} \\ \\ cos B= \frac{(5)^2+(\sqrt{18})^2-(\sqrt{13})^2}{2*5*\sqrt{18} }\\ \\ cosB= \frac{25+18-13}{10\sqrt{18}} =\frac{30}{10\sqrt{18}}=\frac{3}{\sqrt{18}}\\ \\ cosB=\frac{3}{3\sqrt{2}} =\frac{1}{\sqrt{2}}\\ \\ B= cos^-^1(\frac{1}{\sqrt{2}})= 45 degree

So, the measure of angle ABC is 45°

7 0
3 years ago
Read 2 more answers
Why is ln natural logarithm?
klemol [59]
I think its because log e is ln so ln is a natural logarithm.
6 0
3 years ago
Read 2 more answers
What is the completely factored form of 12xy-9x-8y+6
boyakko [2]

Answer: (4y - 3)(3x - 2)

Step-by-step explanation: Factor the polynomial.

Hope this helps! :) ~Zane

5 0
3 years ago
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Simplify 5+2i/6+i
nlexa [21]

Given problem is \frac{(5+2i)}{(6+i)}

To simplify that we need to multiply and divide by conjugate of the denominator

conjugate of denominator 6+i will be 6-i as we just need to change sign of the imaginary part

Now multiply and divide by 6-i


=\frac{(5+2i)}{(6+i)}\cdot\frac{(6-i)}{(6-i)}

=\frac{30-5i+12i-2i^2}{36-6i+6i-i^2}

=\frac{30+7i-2i^2}{36-i^2}

=\frac{30+7i-2\left(-1\right)}{36-\left(-1\right)}

=\frac{30+7i+2}{36+1}

=\frac{32+7i}{37}

=\frac{32}{37}+\frac{7}{37}i

Hence final answer is \frac{32}{37}+\frac{7}{37}i

3 0
2 years ago
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