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3 years ago

The graph shows two lines, A and B.

Mathematics

2 answers:

muminat3 years ago

7 0

Ur 1 solution is where the lines intersect, which is (1,4)

irga5000 [103]3 years ago

5 0

For the system of equations between the two lines, there is only one solution as you can see from the graph.

The point of intersection of the lines is the solution, the intersection point is at:

x = 1, and y =4

(1 , 4).

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Solve the triangle. Round your answers to the nearest tenth.

wel

Answer:

m∠A ≈ 43°

m∠B ≈ 55°

mBC ≈ 20

Step-by-step explanation:

Law of Sines: $\frac{a}{sinA} =\frac{b}{sinB} =\frac{c}{sinC}$

Step 1: Find m∠B

$\frac{29}{sin82} =\frac{24}{sinB}$

Step 2: Solve for ∠B

29sinB = 24sin82°

sinB = 24sin82°/29

B = sin⁻¹(24sin82°/29)

B = 55.038°

Step 3: Find m∠A

180 - (55.038 + 82)

180 - 137.038

m∠A = 42.962°

Step 4: Find BC

$\frac{29}{sin82} =\frac{BC}{sin42.962}$

Step 5: Solve for BC

29sin42.962° = BCsin82°

BC = 29sin42.962°/sin82°

BC = 19.9581

4 0

2 years ago

If cot theta= 4/3, find csc theta

Marrrta [24]

Answer: Option d.

Step-by-step explanation:

The trigonometric identity needed is:

$csc^2\theta=cot^2\theta+1$

Knowing that $cot\theta=\frac{4}{3}$ :

Substitute it into $csc^2\theta=cot^2\theta+1$ :

$csc^2\theta=(\frac{4}{3})^2+1$

Simplify the expression:

$csc^2\theta=(\frac{4}{3})^2+1\\\\csc^2\theta=\frac{16}{9}+1\\\\csc^2\theta=\frac{25}{9}$

Solve for $csc\theta$ . Apply square root at both sides of the expression:

$\sqrt{csc^2\theta}=\±\sqrt{\frac{25}{9}}$

$csc\theta=\frac{5}{3}$

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2 years ago

Ahh I’m dumb, but I need help please!

andreev551 [17]

Answer:

Derik finished first

Step-by-step explanation:

Derik = 35 1/6 = 35.10

other girl = 35.25

3 0

2 years ago

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Help plz:))) I’ll mark you BRAINLIEST !!BRAINLIEST!!

Verdich [7]

Answer:

I think the answer is A

Step-by-step explanation:

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2 years ago

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Solve the following quadratic equation for all values of x in simplest form.

denis-greek [22]

Answer:

$x=\sqrt{7},\:x=-\sqrt{7}$

Step-by-step explanation:

Start by rewriting the equation to standard form $ax^2+bx+c=0$ .

$5(x^2-10)+5=-10,\\5x^2-50+5=-10,\\5x^2-35=0$ .

Factor out the 5, then isolate $x^2$ :

$5x^2-35=0,\\5(x^2-7)=0,\\x^2-7=0,\\x^2=7,\\x=\pm \sqrt{7}.$ .

Therefore, the solutions to this quadratic are:

$x=\sqrt{7},\:x=-\sqrt{7}$

8 0

3 years ago