7²×√144 plss I'm just a kid trying to do my homework and a big exam is coming up and I'm trying my best to study

Your friend claims that it is possible to draw a right triangle where the cosine of either angle (theta) is exactly the same val

avanturin [10]

Answer:

B. No

$Cos \ \theta\neq Cos (90-\theta)\textdegree$

Step-by-step explanation:

-A right angle triangle has two complimentary acute angles and one right angle.

- $\theta$ is usually one of the acute angles and is equivalent to 90º minus it's complimentary acute angle.

-Complimentary angles add up to 90º.

#For complimentary angles:

$Sin \ \theta=Cos \ (90-\theta)\textdegree\\\\Cos \ \theta=Sin(90-\theta)\textdegree\\\\\therefore Cos \ \theta\neq Cos (90-\theta)\textdegree$

The two acute angles cannot have the same Cosine value.

Hence, she's not correct.

5 0

3 years ago

Write an equation in point-slope form of the line that passes through the point (5, 6) with slope 3/5

I am Lyosha [343]

Answer:

(y - 6) = 3/5(x - 5)

Step-by-step explanation:

First, let's look at what point-slope form is: (y - y₁) = m(x - x₁)

Let's plug in the vallues we know.

m is the slope: 3/5

x₁ is 5

y₁ is 6

Therefore:

(y - 6) = 3/5(x - 5)

This is your equation.

Hope this helps!

8 0

3 years ago

Help me pls i have been stuck on this for the past hour

Sedbober [7]

Answer:

-4, -6, -3, -5, -1. The inequality solved for n is n ≥ -6.

Step-by-step explanation:

Substitute all the values in the equation.

n/2 ≥ -3

-10/2 ≥ -3

-5 is not ≥ -3.

n/2 ≥ -3

-7/2 ≥ -3

-3.5 is not ≥ -3.

n/2 ≥ -3

-4/2 ≥ -3

-2 is ≥ -3.

n/2 ≥ -3

-9/2 ≥ -3

-4.5 is not ≥ -3.

n/2 ≥ -3

-6/2 ≥ -3

-3 is ≥ -3.

n/2 ≥ -3

-3/2 ≥ -3

-1.5 is ≥ -3.

n/2 ≥ -3

-8/2 ≥ -3

-4 is not ≥ -3.

n/2 ≥ -3

-5/2 ≥ -3

-2.5 is ≥ -3.

n/2 ≥ -3

-2/2 ≥ -3

-1 is ≥ -3.

To solve the inequality n/2 ≥ -3 for n, do these steps.

n/2 ≥ -3

Multiply by 2.

n ≥ -6.

3 0

3 years ago

Find the equation, (f(x) = a(x - h)2 + k), for a parabola containing point (2, -1) and having (4, -3) as a vertex. What is the s

Nataliya [291]

Answer:

$f(x)=\frac{1}{2}x^2-4x+5$

Step-by-step explanation:

A parabola is written in the form

$f(x)=a((x-h)^2+k)$ (1)

where:

$h$ is the x-coordinate of the vertex of the parabola

$ak$ is the y-coordinate of the vertex of the parabola

$a$ is a scale factor

For the parabola in the problem, we know that the vertex has coordinates (4,-3), so we have:

$h=4$ (2)

$ak=-3$

From this last equation, we get that $a=\frac{-3}{k}$ (3)

Substituting (2) and (3) into (1) we get the new expression:

$f(x)=-\frac{3}{k}((x-4)^2+k) = -\frac{3}{k}(x-4)^2 -3$ (4)

We also know that the parabola contains the point (2,-1), so we can substitute

x = 2

f(x) = -1

Into eq.(4) and find the value of k:

$-1=-\frac{3}{k}(2-4)^2-3\\-1=-\frac{3}{k}\cdot 4 -3\\2=-\frac{12}{k}\\k=-\frac{12}{2}=-6$

So we also get:

$a=-\frac{3}{k}=-\frac{3}{-6}=\frac{1}{2}$

So the equation of the parabola is:

$f(x)=\frac{1}{2}((x-4)^2 -6)$ (5)

Now we want to rewrite it in the standard form, i.e. in the form

$f(x)=ax^2+bx+c$

To do that, we simply rewrite (5) expliciting the various terms, we find:

$f(x)=\frac{1}{2}((x^2-8x+16)-6)=\frac{1}{2}(x^2-8x+10)=\frac{1}{2}x^2-4x+5$

6 0

3 years ago

Expand (2x-5y)⁷ and find fourth form

elena55 [62]

Step-by-step explanation:

By binomial theorem,

T(r+1) = nCr * a^(r+1) * b^r

Term 4 = 7C3 * (2x)^4 * (-5y)^3 = 35 * (16x^4) * (-125y^3) = -70000x^4y^3.

3 0

3 years ago

7²×√144 plss I'm just a kid trying to do my homework and a big exam is coming up and I'm trying my best to study​

7²×√144 plss I'm just a kid trying to do my homework and a big exam is coming up and I'm trying my best to study