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3 years ago

WILL GIVE BRAINLIEST Solve for x

Mathematics

1 answer:

myrzilka [38]3 years ago

8 0

Answer:

Step-by-step explanation:

sin x° = $\frac{17}{19}$

x° = $sin^{-1}$ $\frac{17}{19}$

x° ≈ 63°

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Find the slope using the formula. (3,-2) and (3, 6)

dangina [55]

Answer:

m = undef

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Point (3, -2)

Point (3, 6)

Step 2: Find slope m

Substitute: $m=\frac{6+2}{3-3}$
Add/Subtract: $m=\frac{8}{0}$
Divide: m = undef

When we divide anything by zero, we will get undefined. Therefore, this would be a vertical line.

6 0

3 years ago

(4x-3)÷(2x+3) + (3+2x)÷(3x^2) =2 so what can x not be

Rudik [331]

Remember that if an x=value makes a denominator 0, the entire equation is undefined.

$\frac{4x-3}{2x+3}+\frac{3+2x}{3x^{2}}=2\\\\2x+3=0\\\\2x=-3\\\\x=\frac{-3}{2}\\\\\\3x^{2}=0\\\\x^{2}=0\\\\x=0$

So whenever x = -3/2 or x = 0 the equation is undefined, and that means that x cannot be -3/2 or 0.

3 0

3 years ago

Please help! I’ll give you brainlyest answer!

Mrrafil [7]

Answer:

The answer to your question is the first choicd

3 0

3 years ago

Can someone help me solve this problem by factoring. x^2 + 8x + 16 = 0

Pavlova-9 [17]

Factorization: (x + 4)∧2=0

Solutions based on Factorization: x + 4 = 0 ⇒ X∨1 = X∨2 = -4

8 0

4 years ago

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Which of the functions graphed below has a removable discontinuity? A B C D

PtichkaEL [24]

<h2>Answer with explanation:</h2>

We know that a removable discontinuity occurs when:

The left and the right hand limit of the function exist at a point and are equal but is unequal to the function's value at that point.

Also it is a point on the graph such that it is undefined at that point.

The graph that has a removable discontinuity is attached to the answer.

Since, at x=0 the left hand and the right hand limit of the function exist but the function is not defined at x=0 , since in the graph there is a open circle at x=0 that means that the point is removed from the range.

5 0

3 years ago

Read 2 more answers