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artcher [175]
3 years ago
6

Solve for g: 2g-3h=13

Mathematics
1 answer:
Sladkaya [172]3 years ago
4 0

(3h/2)+(13/2) Move the terms that don't have g to the other side.

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SOMEONE HELP
pantera1 [17]

Answer:

D

Step-by-step explanation:

Given y varies directly as x and z then the equation relating them is

y = kxz ← k is the constant of variation

To find k use the condition y = \frac{8}{3} when x = 1 and z = 4 , then

\frac{8}{3} = k × 1 × 4 = 4k ( divide both sides by 4 )

\frac{2}{3} = k

y = \frac{2}{3} xz ← equation of variation

When x = 6 and z = 3 , then

y = \frac{2}{3} × 6 × 3 = \frac{2}{3} × 18 = 2 × 6 = 12

6 0
3 years ago
Read 2 more answers
HELP ASAP I’ll give brainzless
dem82 [27]
All of them.

1. If you multiply by 2, they will be equal.
2. If you multiply by 5000 they will be equal.
3. If you divide by 3 they will be equal.
3 0
3 years ago
Sarah backstrokes at an average speed of 8 m/s. how long will it take her to complete the race of 200 meters length?
SCORPION-xisa [38]
200 m divided by 8m/s = 0.025
4 0
3 years ago
The temperature in one northern city was -12°F and in a southern city was 57°F. How much warmer
lara [203]

Answer

69°F

Because

12+57=69

3 0
3 years ago
(x+7)/(x^2-49) find the domain. show work
harina [27]
\large\begin{array}{l} \textsf{Find the domain of}\\\\ \mathsf{f(x)=\dfrac{x+7}{x^2-49}}\\\\ \mathsf{f(x)=\dfrac{x+7}{x^2-7^2}}\\\\\\ \textsf{Factor out the denominator using special products:}\\\\ \textsf{(a difference of squares)}\\\\ \mathsf{f(x)=\dfrac{x+7}{(x+7)(x-7)}} \end{array}


\large\begin{array}{l} \textsf{Restrictions for the domain:}\\\\ \bullet~~\textsf{Denominators must not be zero:}\\\\ \mathsf{(x+7)(x-7)\ne 0}\\\\ \begin{array}{rcl} \mathsf{x+7\ne 0}&~\textsf{ and }~&\mathsf{x-7\ne 0}\\\\ \mathsf{x\ne -7}&~\textsf{ and }~&\mathsf{x\ne 7} \end{array} \end{array}


\large\begin{array}{l} \textsf{Therefore, the domain of f is}\\\\ \mathsf{D_f=\{x\in\mathbb{R}:~~x\ne -7~~and~~x\ne 7\}}\\\\\\ \textsf{or using a more compact form}\\\\ \mathsf{D_f=\mathbb{R}\setminus\{-7,\,7\}}\\\\\\ \textsf{or using the interval notation}\\\\ \mathsf{D_f=\left]-\infty,\,-7\right[\,\cup\,\left]7,\,+\infty\right[.} \end{array}


<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2155752


\large\textsf{I hope it helps. :-)}



Tags: <em>function domain real rational factorizing special product interval</em>

</span>
7 0
3 years ago
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