3 years ago

Can someone help me with these ?

Mathematics

1 answer:

I am Lyosha [343]3 years ago

4 0

Multiply width times height times length

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Solve (x+5)+4=16 by first subtracting 4 and then 5. Show your work and justify each step.

Vinil7 [7]

$(x+5)+4=16\ \ \ subtract\ 4 \ and \ 5 \ to\ both\ sides \\\\x+5+4-5-4=16-4-5 \\\\x=7$

3 0

3 years ago

PLZ HELP NNOOOOOOOWW!! 15 points!!

motikmotik

The midpoint of the line having the endpoints $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_{1}+_x_{2}}{2}, \frac{y_{1}+_y_{2}}{2})$
basically average them

so given that the line has the endpoints of (-3,7) and (9,-2)
x₁=-3
y₁=7
x₂=9
y₂=-2

so the midpoint can be found by doing
$(\frac{-3+9}{2}, \frac{7+(-2)}{2})=$
$(\frac{6}{2}, \frac{7-2}{2})=$
$(3, \frac{5}{2})$

the midpoint is $(3,\frac{5}{2})$

4 0

3 years ago

Show by substitution whether the number r is a solution of the corresponding quadratic equation.

Stels [109]

Answer:

It's choice B.

Step-by-step explanation:

When r= -11,

x^2 + 8x - 33 = (-11)^2 - 88 - 33

= 121 - 88 - 33

= 0

- so x = -11 gives a zero value and is a solution to the equation.

3 0

2 years ago

Can someone please solve this

mote1985 [20]

That’s pretty easy you just have to basically y=mx+b

5 0

2 years ago

Which is the equation of a hyperbola centered at the origin with vertex (0, sqrt 12 ) that passes through (2 sqrt 3,6)

mina [271]

Answer:

y-intercept = a +y^2/12 term in it.

Then for y=6, you have

.. 6^2/12 -x^2/b = 1

.. 2 = x^2/b

.. b = x^2/2

If your point is (2√3, 6), then this is

.. b = (2√3)^2/2 = 12/2 = 6

Then the hyperbola's equation is

.. y^2/12 -x^2/6 = 1 . . . . . . . . selection D

Step-by-step explanation:

6 0

3 years ago