All categories

3 years ago

1) What is the equation of a line that is parallel to the line that is given by the equation x = -4?

Mathematics

1 answer:

tester [92]3 years ago

8 0

Not sure its A but use this http://www.mathpapa.com/algebra-calculator.html

You might be interested in

PLEASE HELP ASAP I am kinda stuck on this because my teacher didn’t teach me this

wariber [46]

Answer:

the first option

a = -2

b = 3

Step-by-step explanation:

See photo attached. (:

5 0

3 years ago

What is the sum of 1/9, 2/3,and 5/18? A.19/18 B.8/30 C.4/15 D.12/9

amm1812

1/9 times 2/2=2/18
2/3 times 6/6=12/18

2/18+12/18+5/18=(2+12+5)/18=19/18=1 and 1/8

A is answer

8 0

3 years ago

Pls, help im very confused thanks!

puteri [66]

Answer:

D part

Step-by-step explanation:

3x + 80⁰ + 2x = 180⁰ (By Linear Pair)

5x + 80⁰ = 180⁰

5x = 100⁰

x = 20⁰

Hope it helps :)

4 0

2 years ago

Read 2 more answers

Qual é a frequência de cada gênero de leitura citados?

Karo-lina-s [1.5K]

Answer:

............................

6 0

3 years ago

The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c

Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 $\rm{in^3}$

Step-by-step explanation:

The linearization or linear approximation of a function $f(x)$ is given by:

$f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}$ where $df$ is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Taking $R_0=1.5$ inches and $h=3$ inches and evaluating the partial derivatives we obtain:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

substituting the values and taking $dx=0.1$ and $dh=0.3$ inches we have:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$

Therefore the change in volume is estimated to be 17.20 $\rm{in^3}$

4 0

3 years ago