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gavmur [86]
3 years ago
9

I neeed help asap please

Mathematics
1 answer:
jekas [21]3 years ago
8 0

With what I can help

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Plz helppp meee wallah I swear to god I will mark you as a brainliest and I will give you 20
givi [52]

Answer:

Step-by-step explanation:

Equation for the linear function has been given as,

y = 26 - 4x

For x = 1,

y = 26 - 4(1)

  = 26 - 4

  = 22

For x = 2,

y = 26 - 4(2)

  = 26 - 8

  = 18

For x = 3,

y = 26 - 4(3)

  = 26 - 12

  = 14

For x = 6,

y = 26 - 4(6)

  = 26 - 24

  = 2

Therefore, table for the function will be,

x               y

1              22  

2              18

3              14

6               2

8 0
3 years ago
ASAP—-what is the missing value
olga_2 [115]

Answer:

im sure its 139

Step-by-step explanation:

because you do 128 + 93 which is 221

next, you do 360 (triangle angles) then you do 360-221

which gives you 139!

hope it helped! =)

8 0
3 years ago
Find the cosine of each acute angle
Lemur [1.5K]

Answer:

cos P = 53 degrees

cos Q = 37 degrees

Step-by-step explanation:

cos P = \frac{adjacent}{hypothenuse}

cos P = \frac{24}{40}

cos P = 0.60

Cos-1 of 0.60 is 53.13 (round to the nearest whole number)

cos P = 53 degrees

cos Q = \frac{adjacent}{hypothenuse}

cos Q = \frac{32}{40}

cos Q = 0.80

Cos of 0.80 is 36.86 (round to the nearest whole number)

cos Q = 37 degrees

To check anser:

90 + 53 + 37 = 180

180 = 180

4 0
3 years ago
Cot^2x/cscx-1=1+sinx/sinx
KATRIN_1 [288]
\bf \textit{difference of squares}
\\\\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)
\\\\\\
sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
-------------------------------\\\\
\cfrac{cot^2(x)}{csc(x)-1}=\cfrac{1+sin(x)}{sin(x)}\impliedby \textit{let's do the left-hand-side}

\bf \cfrac{\quad \frac{cos^2(x)}{sin^2(x)}\quad }{\frac{1}{sin(x)}-1}\implies \cfrac{\quad \frac{cos^2(x)}{sin^2(x)}\quad }{\frac{1-sin(x)}{sin(x)}}\implies \cfrac{cos^2(x)}{sin^2(x)}\cdot \cfrac{sin(x)}{1-sin(x)}
\\\\\\
\cfrac{cos^2(x)}{sin(x)}\cdot \cfrac{1}{1-sin(x)}\implies \cfrac{cos^2(x)}{sin(x)[1-sin(x)]}

\bf \cfrac{1-sin^2(x)}{sin(x)[1-sin(x)]}\implies \cfrac{1^2-sin^2(x)}{sin(x)[1-sin(x)]}
\\\\\\
\cfrac{\underline{[1-sin(x)]}~[1+sin(x)]}{sin(x)\underline{[1-sin(x)]}}\implies \cfrac{1+sin(x)}{sin(x)}
5 0
3 years ago
Evaluate the following equation: integral from 3 to infinity: 1/(x-2)^(3/2) dx
Bond [772]

The formula in solving the integral of the infinity of 3 is  ∫3<span>∞?</span>(1<span>)÷((</span>x−2<span><span>)<span><span>(3/</span><span>2)</span></span></span>)</span><span>dx</span>

Substitute the numbers given then solve

limn→inf∫3n(1/((n−2)(3/2))dn

limn→inf[−2/(n−2−−−−−√)−(−2/3−2−−−−√)

=0+2=2

 

Solve for the integral of 2 when 2 is approximate to 0. Transpose 2 from the other side to make it -2

∫∞3(x−2)−3/2dx=(x−2)−1/2−1/2+C

(x−2)−1/2=1x−2−−−−√

0−(3−2)−1/2−1/2=2

<span> </span>

8 0
4 years ago
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