All categories

4 years ago

Can anyone help me with this?

Mathematics

2 answers:

Anna35 [415]4 years ago

6 0

The answer is B the second one

vivado [14]4 years ago

4 0

The answer is the second one

You might be interested in

Someone help me ASAP!!!! I need this ASAP!!!!

qwelly [4]

So to solve it the equation is a squared plus b squared = c squared. So you square a and b. Then do the square root of that and should get c. C is always the largest side. C is opposite from a 90 degree angle . Hope this helps you answer your question.

5 0

3 years ago

Write an equation in point-slope form and in slope-intercept form.

Sever21 [200]

The point-slope form:

$y-y_1=m(x-x_1)$

The slope-intercept form:

$y=mx+b$

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (1, -5) and (-3, -7). Substitute:

$m=\dfrac{7-(-5)}{-3-1}=\dfrac{12}{-4}=-3\\\\y-(-7)=-3(x-(-3))\\\\\boxed{y+7=-3(x+3)}$

Rewrite to the slope-intercept form:

$y+7=-3(x+3)$ <em>use distributive property</em>

$y+7=-3x-9$ <em>subtract 7 from both sides</em>

$\boxed{y=-3x-16}$

3 0

3 years ago

A positive real number is 6 less than another. If the sum of the square of the two numbers is 38, then finds the numbers.

Serggg [28]

Answer:

There seems to be a typo error but still this has solution.

Let the number be = x

A positive real number is 6 less than another. Means the second will be $(x+6)$

The sum of the square of the two numbers is 38.

$x^{2} +(x+6)^{2} =38$

=> $x^{2} +36+12x+x^{2} =38$

=> $2x^{2}+12x+36 =38$

=> $2x^{2}+12x=2$

=> $2x^{2}+12x-2=0$

Taking out 2 common;

=> $x^{2}+6x-1=0$

Solving this quadratic equation, we get;

$x=-3+\sqrt{10}$ and $x=-3-\sqrt{10}$

As positive number is needed, we have $-3+\sqrt{10}$

or x = $\sqrt{10}-3$

x = 0.162277

And other number is 6.162277.

6 0

3 years ago

Read 2 more answers

Use the relationship between the angles in the figure to answer the question.

Anuta_ua [19.1K]

Answer:

37+x+53 = 180

x=90

Step-by-step explanation:

To find x, we need to use the fact that a straight line equals 180 degrees.

37+x+53 = 180

Combining like terms

90+x = 180

Subtract 90 from each side

90-90+x = 180-90

x = 90

4 0

3 years ago

Read 2 more answers

Suppose that receiving stations X, Y, and Z are located on a coordinate plane at the points (4,5), (-6,-6), and (-14,2

Lilit [14]

Answer:

(-2, -3)

Step-by-step explanation:

A careful graph shows the point (-2, -3) is at the intersection of the circles whose radii are the given distances from the receiving stations.

_____

The simultaneous equations for the circles can be solved algebraically.

The epicenter is 10 units from X, so lies on the circle ...

(x -4)^2 +(y -5)^2 = 10^2

x^2 -8x +16 +y^2 -10y +25 = 100

x^2 +y^2 -8x -10y = 59

The epicenter is 5 units from Y, so lies on the circle ...

(x +6)^2 +(y +6)^2 = 5^2

x^2 +12x +36 +y^2 +12y +36 = 25

x^2 +y^2 +12x +12y = -47

The epicenter is 13 units from Z, so lies on the circle ...

(x +14)^2 +(y -2)^2 = 13^2

x^2 +28x +196 +y^2 -4y +4 = 169

x^2 +y^2 +28x -4y = -31

Subtracting the second equation from each of the other two, we get ...

(x^2 +y^2 -8x -10y) -(x^2 +y^2 +12x +12y) = (59) -(-47)

-20x -22y = 106 . . . . eq1 -eq2

(x^2 +y^2 +28x -4y) -(x^2 +y^2 +12x +12y) = (-31) -(-47)

16x -16y = 16 . . . . . . . .eq3 -eq2

These simultaneous linear equations can be solved a variety of ways. We might use substitution:

x = y+1 . . . . . from eq3 -eq2 divided by 16

10(y +1) +11y = -53 . . . . . from eq1 -eq2 divided by -2

21y = -63 . . . . . . . . . . . . simplify, subtract 10

y = -3

x = y+1 = -2

The epicenter is located at (x, y) = (-2, -3).

8 0

3 years ago