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3 years ago

What is the reasoning for this statement?

Mathematics

1 answer:

Dahasolnce [82]3 years ago

4 0

Answer:

there is a black photo

Step-by-step explanation:

make sure to check properly

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What is the equation of a line that is parallel to the line 2x + 5y = 10 and passes through the point (-5, 1)? Check all

Lilit [14]

Answer:

B, E

Step-by-step explanation:

The equation of that line in slope intercept form is y = -2/5x + 2

The slope of a line parallel to that line will be the same as its slope, so -2/5.

To find the y-intercept of a line that passes through the point (-5, 1) with that slope, you will have to plug in the x and y values of that point into what you know of the equation of the line.

y = -2/5x + b

1 = -2/5(-5) + b

1 = 2 + b

-1 = b

From this, you can construct an equation.

y = -2/5x - 1

However, this is not an answer choice.

It cannot be A because this line does not have a slope of -1.

2x + 5y = 15

5y = -2x - 5

y = -2/5x - 1

It can be B because this line is the same as the equation we came up with.

It cannot be C because this line does not have a slope of -1 or a y-intercept of -3.

2x + 5y = -15

5y = -2x - 15

y = -2/5x - 3

It cannot be D because this line does not have a y-intercept of -3.

y - 1 = -2/5(x + 5)

y - 1 = -2/5x - 2

y = -2/5x - 1

It can be E because this equation matches the one we came up with.

5 0

3 years ago

Read 2 more answers

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

Is this answer true. X is 6.<br><br> X+14>5

Anna35 [415]

Answer:

x is a solution

Step-by-step explanation:

X+14>5

Subtract 14 from each side

X+14-14>5-14

x > -9

Substitute the value in

6 >-9

This is true so x is a solution

7 0

3 years ago

Read 2 more answers

PLEASE HELP ME AND SHOW YOUR WORK

Tcecarenko [31]

Answer:

Step-by-step explanation:

Hope the picture helps!

3 0

3 years ago

You might need:

Korolek [52]

Answer:

Step-by-step explanation:

so I can't see the picture or what the "Shaded region " is.. but I can guess that it's the part of the 3x3 rectangle with the circle removed.. right?

if that's the case... then

3*3 = 9 $cm^{2}$

9 - area of circle = shaded region

area of circle = pi * $r^{2}$

pi * 1 cm = 3.1415 $cm^{2}$

9 $cm^{2}$ - 3.1415 $cm^{2}$ = 5.8585 $cm^{2}$

5.86 $cm^{2}$

I don't see this as a choice of answers so maybe that's not what the shaded region is. any way.. hope it helps

7 0

3 years ago