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3 years ago

What is the value of x in sin 29° = cosx? 29 degrees 61 degrees 1 degree 151 degrees

Mathematics

1 answer:

VladimirAG [237]3 years ago

4 0

Answer:

61 degrees

Step-by-step explanation:

We need to take the inverse cosine of both sides as follows:

cos^-1(sin(29))=cos^-1(cos(x))

61=x

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Determine the volume of the composite figure

madreJ [45]

1.8903
because if you do the math its easy

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3 years ago

It’s confusing, would the answer be square root 7 over 4?

Colt1911 [192]

Answer:

$\tan U = 1$

Step-by-step explanation:

$\text{Apply Pythagorean theorem,}\\\\~~~~~~\text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2\\\\\implies UT^2 = UV^2 +VT^2\\\\\implies UV^2 = UT^2 -VT^2\\\\\implies UV^2 = \left(7\sqrt 2 \right)^2 - 7^2\\\\\implies UV^2 = 49(2)-49\\\\\implies UV^2 = 49\\\\\implies UV = \sqrt{49} \\\\ \implies UV = 7$

$\text{Now,}\\\\~~~~~~~~\tan \theta = \dfrac{\text{Perpendicular}}{\text{Base}}\\\\\\\implies \tan U = \dfrac{7}{7}\\\\\\\implies \tan U = 1$

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2 years ago

Ronald bikes 6.9 miles each day. How far has Ronald biked in seven days

FromTheMoon [43]

Answer: 48.3 6.9 times seven

8 0

3 years ago

Read 2 more answers

Find t12 term in an arithmetic progression having t3 = 10 and t10 = −4 .

Marat540 [252]

Answer:

t12= -8

Step-by-step explanation:

a+2d=10 (1)

t10

a+9d=-4. (2)

From (1)

a=10-2d. (3)

Sub into (3) into equ (2)

a+9d=-4

10-2d+9d=-4

10+7d=-4

7d=-4-10

7d=-14

Divide both sides by 7

d=-14/7

d= -2

10=a+2d

10=a+2(-2)

10=a-4

10+4=a

14=a

a=14

t12=a+11d

= 14+11(-2)

=14-22

= -8

t12= -8

6 0

3 years ago

Given that 'n' is a natural number. Prove that the equation below is true using mathematical induction.

LenaWriter [7]

<h3>To ProvE :- </h3>

1 + 3 + 5 + ..... + (2n - 1) = n²

Method :-

If P(n) is a statement such that ,

P(n) is true for n = 1
P(n) is true for n = k + 1 , when it's true for n = k ( k is a natural number ) , then the statement is true for all natural numbers .

$\sf\to \textsf{ Let P(n) : 1 + 3 + 5 + $\dots$ +(2n-1) = n$^{\sf 2}$ }$

Step 1 : Put n = 1 :-

$\sf\longrightarrow LHS = \boxed{\sf 1 } \\$

$\sf\longrightarrow RHS = n^2 = 1^2 = \boxed{\sf 1 }$

Step 2 : Assume that P(n) is true for n = k :-

$\sf\longrightarrow 1 + 3 + 5 + \dots + (2k - 1 ) = k^2$

Add (2k +1) to both sides .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1)=k^2+(2k+1)$

RHS is in the form of ( a + b)² = a²+b²+2ab .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1)= (k +1)^2$

Adding and subtracting 1 to LHS .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1) + 1 -1 = (k +1)^2 \\$

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+2) - 1 = (k +1)^2$

Take out 2 as common .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+\{2(k+1)-1\}= (k +1)^2$

P(n) is true for n = k + 1 .

Hence by the principal of Mathematical Induction we can say that P(n) is true for all natural numbers 'n' .

**Edits are welcomed**

8 0

3 years ago

Read 2 more answers