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Maslowich
3 years ago
15

Hi guys or eveyone?????

Mathematics
2 answers:
jasenka [17]3 years ago
7 0
Hellooooooo lols thanks for the free points
weqwewe [10]3 years ago
5 0
Hiiiiiiiiiii lollllllll
You might be interested in
What is x-intercept of y=-3x+5
Yuri [45]

Answer:

To find x intercept, u graph it

the x intercept is 1.67

Step-by-step explanation:

6 0
3 years ago
Which of the following is the right algerbraic expression??
Sliva [168]

Answer:

A

Step-by-step explanation:

Area of rectangle = l x w
A = ac

Area of circle = πr^2

r = d/2 = b/2

b = semicircle = half of normal circle

A = π\frac{b^{2} }{2} + ac

4 0
2 years ago
Does anyone know how to solve the equation ln 2 - ln(x - 8) = 4, giving nearly 80 points
RoseWind [281]

Answer:

x= 8.1353

x= 8.135 (rounded to the nearest tenth-thousandths)

x= 8.14 (rounded to the nearest thousandths)

x= 8.1 (rounded to the nearest tenth)

Step-by-step explanation:

<u><em>Note I am not 100% sure with my answer</em></u>

2 − ln (x − 8)= 4

−ln (x − 8) + 2= 4

−ln (x − 8) + 2 + −2= 4 + −2

−ln (x − 8)= 2

−ln (x − 8)/−1= 2/−1

ln (x − 8)= −2

<solve for the logarithm (ln)>

ln (x − 8)= −2

e^ln (x − 8)= e^−2

x − 8= e^−2

x − 8= 0.1353

x − 8 + 8= 0.1353 + 8

x= 8.1353

5 0
3 years ago
WHATS THE ANSWER PLEASE HELP AND EXPLAIN IT TO MEEEEEE
Veseljchak [2.6K]
Can I have more explanation please? :)
4 0
3 years ago
Prove that: (Sec A- cosec A)(1+ tan A+cot A) = tan A× sec A - cot A × cosec A
mash [69]

Answer:

Step-by-step explanation:

(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA

we take the LHS so here goes,

(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA\\(\frac{1}{cosA} -\frac{1}{sinA})(1+\frac{sinA}{cosA}+\frac{cosA}{sinA})\\\\(\frac{sinA-cosA}{sinAcosA})(\frac{sinAcosA+sin^2A+cos^2A}{sinAcosA})\\

since , sin^2A+cos^2A=1

the identity becomes,

(\frac{sinA-cosA}{sinAcosA})(\frac{1+sinAcosA}{sinAcosA})\\\\(\frac{sinA+sin^2AcosA-cosA-cos^2AsinA}{sin^2Acos^2A})\\\\

now, we know,

sin^2A=1-cos^2A and cos^2A=1-sin^2A

the identity becomes,

(\frac{sinA+(1-cos^2A)cosA-cosA-(1-sin^2A)sinA}{sin^2Acos^2A} )\\\\

(\frac{sinA+cosA-cos^3A-cosA-sinA+sin^3A}{sin^2Acos^2A})

sin A and cos A cancel out it becomes zero

\frac{sin^3A-cos^3A}{sin^2Acos^2A} \\\\

Splitting the denominator the identity becomes

\frac{sin^3A}{sin^2Acos^2A}-\frac{cos^3A}{sin^2Acos^2A}  \\\\\frac{sinA}{cos^2A} - \frac{cosA}{sin^2A} \\\\\frac{sinA}{cosA}(\frac{1}{cosA})-\frac{cosA}{sinA}(\frac{1}{sinA})\\\\

Hence,

tanAsecA-cotAcosecA

3 0
3 years ago
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