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3 years ago

−4+25=−4+25 solve for x show working

Mathematics

2 answers:

fredd [130]3 years ago

6 0

Answer:

you can solve this anymore....

Step-by-step explanation:

pychu [463]3 years ago

4 0

Answer:

Its already solved, -4+25 does equal -4+25

Step-by-step explanation:

You might be interested in

Your family wants to purchase a new laptop with

zloy xaker [14]

Answer:

The laptop will not fit in the case.

Step-by-step explanation:

We need to find the width of the laptop to know if it fits in the case. We have:

For the laptop:

h: is the height of the laptop = 10 inch

d: is the widescreen of the laptop (diagonal) = 17 inch

w: is the width of the laptop =?

For the laptop case:

H: is the height of the laptop case = 10.5 inch

W: is the width of the laptop case = 13 inch

So, the width of the laptop can be found by using Pitagoras:

$d^{2} = w^{2} + h^{2}$

$w = \sqrt{(17 inch)^{2} - (10 inch)^{2}} = 13.7 inch$

Hence, the laptop is 13.7 inches wide.

Therefore, since the laptop is wider than the case, the laptop will not fit in the case.

I hope it helps you!

8 0

2 years ago

Find the length of AB

aniked [119]

FIRST QUESTION
Given points are:
A(2, -4) and B(6, 2)
Now, Use the distance formula.
distance formula = $\sqrt{ (x_{2}- x_{1})^{2} + ( y_{2} - y_{1} )^{2} }$

Now, plug the values into the formula, So,
distance = $\sqrt{ (6- 2)^{2} + ( 2 - (-4))^{2} }$

= $\sqrt{ (6- 2)^{2} + ( 2 +4))^{2} }$

= $\sqrt{ (4)^{2} + ( 6))^{2} }$

= $\sqrt{ 16+36}$

= $\sqrt{52}$

= $2 \sqrt{13}$

So, the length of AB is $2 \sqrt{13}$ .

THIRD QUESTION
Two points given are:
A(3, -2) and B(1, 1)
Also given that B is the midpoint of AC.

Let, the co-ordinates of C be C(a, b).
Now, using midpoint formula,
Midpoint = $(\frac{ x_{1}+ x_{2} }{2} , \frac{ y_{1}+ y_{2} }{2} )$

$(1, 1)=(\frac{ 3+ a }{2} , \frac{ -2+b }{2} )$

Now, equaling the ordered pair, we have,

$1=\frac{ 3+ a }{2}$ .............equation (1)

$1=\frac{ -2+b }{2}$ ................equation (2)

Now, taking equation (1)
$1=\frac{ 3+ a }{2}$

$1*2=3+a$

$2-3=a$

$a=-1$

Now, taking equation (2)
$1=\frac{ -2+b }{2}$

$1*2=-2+b$

$2+2=b$

$b=4$

So, the co ordinates of C are (a, b) which is (-1 , 4)


SECOND QUESTION:
Given equations are:
2x + 3y = 14.....................equation (1)
-4x + 2y = 4 .....................equation (2)
Taking equation (2)
-4x + 2y = 4
2y = 4 + 4x
y = (4 + 4x) / 2
y = 2 + 2x .......................equation (3)
Now, Taking equation (1)
2x + 3y = 14
Substituting the value of y from equation (3), we get,
2x + 3(2 + 2x) = 14
2x + 6 + 6x = 14
8x = 14 - 6
x = (14 - 6) / 8
x = 1

Taking equation (3)
y = 2 + 2x
Now, substituting the value of x in equation (3), we get,
y= 2 + 2 (1)
y = 2 + 2
y = 4

So, x=1 and y=4

6 0

3 years ago

What function type can have vertical asymptotes?

Sonja [21]

Rational functions of the form
f(x)= polynomial in x / another polynomial in x

See example plot attached.