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2 years ago

Can someone help me on this?

Mathematics

1 answer:

Mnenie [13.5K]2 years ago

5 0

Answer:

Step-by-step explanation:

We know that cosx = sin( $\pi$ - x)

We can apply it here:

sin(55°) = cosx

cos( $\pi$ - 55°) = cosx

Now we apply the arccos to both sides:

x = $\pi$ - 55° + 2k $\pi$ or x = - $\pi$ - 55° + 2k $\pi$

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Evaluate to one decimal place

ANTONII [103]

Answer:

<u>The correct answer is C. 1.6</u>

Step-by-step explanation:

Let's recall that e or Euler's number is a mathematical constant that is the base of the natural logarithm, in other words, the unique number whose natural logarithm is equal to one. The value of e is approximately 2.71828.

Now, replacing e, we have:

2.71828∧(1/2) = √2.71828 = 1.6 (Rounding to one decimal place)

<u>The correct answer is C. 1.6</u>

3 0

4 years ago

Help me pleaseeee I need it

Marta_Voda [28]

4(2x5 + 3) - 4(2) + 2(5).

4(13) - 18.

52-18 = 34.

4 0

3 years ago

Read 2 more answers

Find the discriminant for each quadratic function below. Then list the number of zeros/solutions for each quadratic function.

Hunter-Best [27]

Answer:

Step-by-step explanation:

The discriminant is found in

$b^2-4ac$ which is part of the quadratic formula. In

$4x^2+10x-16=0$ , a = 4, b = 10, c = -16. Filling in for the discriminant:

$10^2-4(4)(-16)=100+256=356$

It's positive so we know we have real roots; it's not a perfect square, so our solutions are 2 complex rational.

In $-16x^2-7x+3=0$ , a = -16, b = -7, c = 3. Filling in for the discriminant:

$(-7)^2-4(-16)(3)=49+192=241$

It's positive and not a perfect square so the 2 solutions are complex rational

5 0

3 years ago

Help pleaseee i don’t understand thisss

kotegsom [21]

Answer:

The answer is;

box A:3/2

box B:2/3

6 0

3 years ago

Find the zeros (x-3)(x+5)=0

maxonik [38]

The product is equal zero if one of the factors is equal zero.Therefore:
$(x-3)(x+5)=0\iff x-3=0\ \vee\ x+5=0\\\\\boxed{x=3\ \vee\ x=-5}$

3 0

3 years ago