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2 years ago

Solve for x 5x 2^3x = 21

Mathematics

1 answer:

GenaCL600 [577]2 years ago

5 0

If u meant to put, 5x multipled by 2^3x=21 the answer is x= 0.798

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Why isn't 20 over 100 proportional

Rina8888 [55]

It isn't proportional because it is not reduced to the lowest terms

Instead 20 over 100 = 1 over 5

Hope this helps!

6 0

3 years ago

I hope someone can help me here please don't answer nonsense!

Karolina [17]

Answer:

30 degrees

Step-by-step explanation:

Angle B is an inscribed angle -- its vertex is on the circle and it intersects the circle in two points, A and C.

The measure of an inscribed angle is always one half the measure of the intercepted arc. The intercepted arc has measure 60 degrees, so the measure of angle B is 30 degrees.

8 0

2 years ago

Part 4: Identify the vertex, focus and directrix of each. Then sketch the graph.

Nezavi [6.7K]

Answer: 1) Vertex: (6, -2) Focus: (6, -7/4) Directrix: y = -9/4

2) Vertex: (-2, -1) Focus: (-7/4, -1) Directrix: x = -9/4

Step-by-step explanation:

Rewrite the equation in vertex format y = a(x - h)² + k or x = a(y - k)² + h by completing the square. Divide the b-value by 2 and square it - add that value to both sides of the equation.

(h, k) is the vertex
p is the distance from the vertex to the focus
-p is the distance from the vertex to the directrix

$\bullet\quad a=\dfrac{1}{4p}$

1) y = x² - 12x + 34

$y-34=x^2-12x\\\\y-34+\bigg(\dfrac{-12}{2}\bigg)^2=x^2-12x+\bigg(\dfrac{-12}{2}\bigg)^2\\\\\\y-34+36=(x-6)^2\\\\y+2=(x-6)^2\\\\y=(x-6)^2-2\qquad \rightarrow \qquad a=1\quad (h,k)=(6,-2)\\$

$a=\dfrac{1}{4p}\quad \rightarrow \quad 1=\dfrac{1}{4p}\quad \rightarrow \quad p=\dfrac{1}{4}\\\\\\\text{Focus = Vertex + p}\\.\qquad \quad =\dfrac{-8}{4}+\dfrac{1}{4}\\\\.\qquad \quad =-\dfrac{7}{4}\quad \rightarrow \quad\text{Focus}=\bigg(6,-\dfrac{7}{4}\bigg)\\\\\\\text{Directrix: y= Vertex - p}\\.\qquad \qquad y=\dfrac{-8}{4}-\dfrac{1}{4}\\\\.\qquad \qquad y=-\dfrac{9}{4}$

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2) x = y² + 2y - 1

$x+1=y^2+2y\\\\x+1+\bigg(\dfrac{2}{2}\bigg)^2=y^2+2y+\bigg(\dfrac{2}{2}\bigg)^2\\\\\\x+1+1=(y+1)^2\\\\x+2=(y+1)^2\\\\x=(y+1)^2-2\qquad \rightarrow \qquad a=1\quad (h,k)=(-2,-1)\\$

$a=\dfrac{1}{4p}\quad \rightarrow \quad 1=\dfrac{1}{4p}\quad \rightarrow \quad p=\dfrac{1}{4}\\\\\\\text{Focus = Vertex + p}\\.\qquad \quad =\dfrac{-8}{4}+\dfrac{1}{4}\\\\.\qquad \quad =-\dfrac{7}{4}\quad \rightarrow \quad\text{Focus}=\bigg(-\dfrac{7}{4},-1\bigg)\\\\\\\text{Directrix: x= Vertex - p}\\.\qquad \qquad y=\dfrac{-8}{4}-\dfrac{1}{4}\\\\.\qquad \qquad x=-\dfrac{9}{4}$

4 0

3 years ago

Please help!!!

kifflom [539]

Answer: If I am correct, I do believe that A is equal to F, meaning that the angle of A, which is 100°, is also the angle of F.

Step-by-step explanation:

Sorry if I am wrong I'm new at this.

8 0

3 years ago

Read 2 more answers

Use the quadratic formula to solve the equation. -2x^2-x+7=0.

Furkat [3]

-2x^2-x+7=0
Variable with the highest degree's (exponent) constant, -2 is a, next variable's constant, -1 is b, the constant or number without a variable, 7 is c

using substitution put the numbers into the formula
(-b±√(b^(2)-4ac))/(a^(2))

(-(-1)±√((-1)^(2)-4(-2)(7))/((-2)^(2)) simplify

(1±√(1+56))/4

1±√(57)/4 is your answer

8 0

3 years ago