All categories

3 years ago

3(g + 4) = 9 Need help

Mathematics

2 answers:

andrezito [222]3 years ago

7 0

Answer:

g=-1

Step-by-step explanation:

3(g+4)=9

Divide both sides by 3.

g+4= 9/3

Divide 9 by 3 to get 3.

g+4=3

Subtract 4 from both sides.

g=3−4

Subtract 4 from 3 to get −1.

g=−1

Leona [35]3 years ago

7 0

Answer:

3(g+4)=9

or, 3(g+4)/3=9/3 [Dividing both sides by 3.]

or, g+4=3

or, g+4-4=3-4 [Subtracting 4 from both sides.]

So, g=-1

You might be interested in

With regard to mathematics, what is an accepted statement of fact that is used to prove other statements

spayn [35]

In mathematics, a proof is a deductive agruement for a mathematical statement.In the argument,other prevoiusly establised statements, such as theroms, an be used. In principle, a proof can be traced back to self-evident or assumed statements, known as axioms, along with accepted rules of inference.

5 0

3 years ago

Help!! Which of the following best represents Z1 • Z2 select all that apply

AURORKA [14]

$z_1=\sqrt{3}\left(cos\:\frac{\pi }{4}+i\:sin\:\frac{\pi }{4}\right)$

$z_2=\sqrt{6}\left(cos\:\frac{3\pi \:}{4}+i\:sin\:\frac{3\pi \:}{4}\right)$

$\cos \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}$

$\sin \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}$

$\cos \left(\frac{3\pi }{4}\right)=-\frac{\sqrt{2}}{2}$

$\sin \left(\frac{3\pi }{4}\right)=\frac{\sqrt{2}}{2}\\$

$Z_1*Z_2=\sqrt{3}\left(cos\:\frac{\pi }{4}+i\:sin\:\frac{\pi }{4}\right)\cdot \sqrt{6}\left(cos\:\frac{3\pi \:}{4}+i\:sin\:\frac{3\pi \:}{4}\right)$

$=\sqrt{3}\left(\frac{\sqrt{2}}{2}+\:i\frac{\sqrt{2}}{2}\right)\cdot \sqrt{6}\left(\frac{-\sqrt{2}}{2}+\:i\frac{\sqrt{2}}{2}\right)$

On simplifying, we get

$Z_1* Z_2 =-3\sqrt{2}$

<h2>Therefore, correct option is 1st option.</h2>

3 0

3 years ago

Plssssss help fast!!!!Launching From a Tower

aleksandrvk [35]

Answer:

A. h = 64

B. t = 4 -> 4 seconds

C. t = 1.5 -> 1.5 seconds

D. maximum height is 100 ft

E. The domain that makes sense for the function in this context is t

is any positive real number since time can not be negative.

Step-by-step explanation:

h = -16t2 + vt + 64

A. What tower platform height was the projectile launched from?

when the projectile was not launched, t = 0

h = -16(0)^2 + v(0) + 64 = 64

B. How long was the projectile in the air?

if the projectile lands, its height = 0 so substitute 0 for h

0 = -16(t)^2 + 48(t) + 64

= -16(t^2 - 3t - 4)

= -16(t - 4)(t + 1)

t = 4 or t = -1

Since time can not be negative, t = -1 can not be the answer. Therefore, the projectile lands when t = 4 or 4 seconds.

C. When did it reach its maximum height?

maximum -> t=-b/2a where in -16(t)^2 + 48(t) + 64, b = 48 and a = -16

t = -48/-32 = 1.5

D. What was its maximum height?

plug t = 1.5 into -16(t)^2 + 48(t) + 64

-16(1.5)^2 + 48(1.5) + 64 = -36 + 72 + 64 = 100

E. The domain that makes sense for the function in this context is t

is any positive real number since time can not be negative.

6 0

2 years ago

Determine the rate of change between the points (-1,-1) and (1,-1).

AVprozaik [17]

Answer:

$\displaystyle m = 0$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinate Planes

Coordinates (x, y)
Slope Formula: $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$

Step-by-step explanation:

*Note:

Rate of change is slope.

Step 1: Define

Identify.

Point (-1, -1)

Point (1, -1)

Step 2: Find slope m

Simply plug in the 2 coordinates into the slope formula to find slope m.

Substitute in points [Slope Formula]: $\displaystyle m = \frac{-1 + 1}{1 + 1}$
[Order of Operations] Simplify: $\displaystyle m = \frac{0}{2}$
Simplify: $\displaystyle m = 0$

4 0

3 years ago

Express this number in standard form. 2.456 times 10^11

Varvara68 [4.7K]

Its 2.456*10=11

hope this helps and please rate below and also i am great with helping people who are going through a lot like family or school or friendship problems so if u want help or tips about those things or even depression feel free to ask me through message and i will see what i can do to help you. :)

5 0

3 years ago

Read 2 more answers