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3 years ago

Can you pls help me with this plsss

Mathematics

1 answer:

Gelneren [198K]3 years ago

8 0

Sure! So you need to make an addition problem, for example, 200+25; that equals 225, right? Now you need to subtract from 225. For example, 225-10= 215. Hope I helped!

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Find all values of the angle θ (in radians, with 0 ≤ θ < 2π) for which the matrix a = cos θ −sin θ sin θ cos θ has real eigen

harina [27]

The matrix

$A=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$

has eigenvalues $\lambda$ such that

$\det(A-\lambda I)=\begin{vmatrix}\cos\theta-\lambda&-\sin\theta\\\sin\theta&\cos\theta-\lambda\end{vmatrix}=0$

$(\cos\theta-\lambda)^2+\sin^2\theta=0$

$(\cos\theta-\lambda)^2=-\sin^2\theta$

$\cos\theta-\lambda=\pm\sqrt{-\sin^2\theta}$

$\lambda=\cos\theta\pm\sqrt{-\sin^2\theta}$

$\sin^2\theta\ge0$ for all values of $\theta$ , so we need to have $\sin\theta=0$ in order for $\lambda$ to be real-valued. This happens for

$\sin\theta=0\implies\theta=n\pi$

where $n$ is any integer, and over the given interval we have $\theta=0$ and $\theta=\pi$ .

4 0

3 years ago

Lucy wants to plant 16 roses, 24 sunflowers, and 32 lilies in her garden. She wants to plant only one type of flower in each row

uranmaximum [27]

Answer:

8 by 8

Step-by-step explanation:

well the GCF of all those are 8 and then adding 16+24+32=64

8 goes into 64 8 times so it would be 8 by 8

5 0

3 years ago

Read 2 more answers

Slove The System of linear equations by graphing. Y=-x+4. 4=2x-8

saveliy_v [14]

Answer:

y= -x+4

4= 2x-8

5 0

3 years ago

It is known that the straight line l is tangent to the circle x^2+y^2=4 at a point on the x-axis, and intersects with the straig

Talja [164]

Step-by-step explanation:

The general equation of a circle is

$(x \ - \ h)^2 \ + \ (y \ - \ k)^2 \ = \ r^{2}$ ,

where h and k forms the coordinates of the centre of the circle.

When the circle has a centre at the origin, the equation reduces into

. $x^2 \ + \ y^2 \ = r^2$ .

Now, we are interested in solving for the x-intercepts (the x-coordinates when the circle intersects the x-axis), of the circle

$x^2 \ + \ y^2 \ = \ 4$ .

Thus,

$x^2 \ + \ (0)^2 \ = \ 4 \\ \\ \\ \-\hspace{1.3cm} x^{2} \ = \ 4 \\ \\ \\ \-\hspace{1.4cm} x \ = \ \pm \ 2$ .

Geometrically speaking, the tangent to the circle at the point defined by one of the x-intercepts of the circle is actually a vertical line, more specifically the lines $x \ = \ \pm \ 2$ .

First and foremost, for the vertical line $x \ = \ 2$ , it intersects the straight line $y \ = \ \displaystyle\frac{1}{2}x$ , giving the y-coordinate for point P,

$y \ = \ \displaystyle\frac{1}{2}(2) \\ \\ \\ y \ = \ 1$ .

Hence, the coordinates of point P are $(1, \ 1)$ .

However, since there are no boundaries given in the question and a circle is symmetrical about its centre, thus, point P also exists when the vertical line $x \ = \ -2$ and interdects the straight line $y \ = \ \displaystyle\frac{1}{2}x$ .

$y \ = \ \displaystyle\frac{1}{2}(-2) \\ \\ \\ y \ = \ -1$ .

Therefore, the coordinates of point P are also $(1, \ -1)$ .

8 0

2 years ago

How do I rewrite In 7=y as an exponential equation? I kind of forgot how to do these

Marina CMI [18]

Answer:

$e^y=7$

Step-by-step explanation:

$ln(7)=y$

$e^{ln(7)}=e^y$

$7=e^y$

Raising e to both sides cancels out the natural logarithm on the left side and we have our exponential form. This works with other bases (recall that the natural logarithm, ln, has base e)

7 0

3 years ago